Topology lectures

ℝ→Metric spaces→convergence, closeness, often different metrics give the same result→Topological space

Advantages

Keep only what you care about

Abstraction→Simplification

More general

Relies on spacial intuition

Modern geometry

Manifolds in higher dimensions: When do they have the same shape?

Fundamental notion: Homeomorphism

Last section Introduction to manifolds

Classify 2-dimensional manifolds

Definition 1. A topological space $(X, 𝒯)$ consists of a non-empty set $X$ together with a family $𝒯$ of subsets of $X$ satisfying:

(T1) $X, ∅∈𝒯$;

(T2) $U, V∈𝒯⇒ U ∩ V∈𝒯$;

(T3) $U_i∈𝒯$ for all $i∈I$ (where $I$ is some indexing set) $⇒ ⋃_{i∈I}U_i∈𝒯$.

The family $𝒯$ is called a topology for $X$. The sets in $𝒯$ are called the open sets of $X$. When $𝒯$ is given we say that $X$ is a topological space.

Remark 2. If $U_1, U_2, \ldots, U_n∈𝒯$ then $U_1 ∩ U_2 ∩ \ldots ∩ U_n∈𝒯$

Proof. By induction□

In general $\bigcap_{i∈ℐ}U_i \not∈𝒯$ if $U_i∈𝒯$.

1Metric spaces

Definition 3. $(X, d)$ is a metric space, $r > 0$, $x∈X$.

The ball of radius $r$ and center $x$ is the set $\{y∈X : d(x, y) < r \}$.

Definition 4. Let $(X, d)$ be metric space. We say that $U⊆X$ is open if $∀ x∈U$ there is $ε_x > 0$ such that $B (x, ε_x)⊆U$.

Proposition 5. The open sets of $X$ form a topology, denoted $𝒯_d$.

Proof. (T1) $∅, X∈𝒯_d$

(T2) $U, V∈𝒯, x∈U ∩ V, B (x, ε_1)⊆U, B (x,ε_2)⊆V ⇒ B (x, ε)⊆U ∩ V,ε=\min (ε_1, ε_2)$.

(T3) Clear□

Remark 6. The open ball $B (x, r)$ is open. Proof: If $y∈B (x, r), d (x, y)=r_1, ε=\frac{r - r_1}{2},$ then $B (y, ε)⊆B (x, r)$.

2More example of topological spaces

$X≠∅$, $𝒯$ all subsets of $X$ (Discrete space)
$X≠∅,$ $𝒯=\{∅, X \}$ (Indiscrete space)
$X=\{1, 2, 3 \}$, $𝒯=\{∅, \{1 \}, \{1, 2 \}, \{1, 2, 3 \}\}$
$X$ infinite, $U$ is open if $U=∅$ or $X∖U$ is finite (Cofinite topology)
Check: T1, T2, T3
(T2) $X∖(U ∩ V)=\underbrace{(X∖U)}_\text{finite}∪\underbrace{(X∖V)}_{\text{finite}}$

Definition 7. A topological space $(X, 𝒯)$ is metrizable if there is a metric $d$ on $X$ such that $𝒯=𝒯_d$.

Definition 8. Two metrics $d_1, d_2$ on $X$ are called topologically equivalent if $𝒯_{d_1}=𝒯_{d_2}$.

Example 9. $d_1, d_2, d_∞$ are topologically equivalent.

$𝒯_{d_1}=𝒯_{d_2}=𝒯_{d_∞}$ is called the standard topology of $ℝ^n$.

Example 10. Let $X≠∅$, define $d : X×X→ℝ$ by $d (x, y)=\begin{cases}1 & \text{ if }x≠ y\\ 0 & \text{ if }x=y\end{cases}$

$𝒯_d$ is the discrete topology, $B \left( x, \frac{1}{2}\right)=\{x \}$, so $∀ U⊆X$, $U$ is open. Discrete topology is metrizable.

Lemma 11. Let $(X, 𝒯)$ be a topological space, then $U⊆X$ is open iff $∀ x∈U, ∃U_x⊆U$ such that $x∈U_x$ and $U_x$ is open.

Proof. ($⇒$) If $U$ is open, just take $U_x=U ∀ x∈U$.

($⇐$) We have $U=⋃_{x∈U}U_x$ and $U_x$ is open, so $U$ is open.□

Definition 12. Given two topologies $𝒯_1, 𝒯_2$ on $X$, we say that $𝒯_1$ is coarser than $𝒯_2$ if $𝒯_1⊆𝒯_2$.

Example 13. If $(X, 𝒯)$ is topological space, then $𝒯$ is coarser than the discrete topology on $X$. The indiscrete topology is coarser than $𝒯$.

Definition 14. Let $(X, 𝒯)$ be topological space, then $K⊆X$ is closed if $X∖K$ is open.

Remark 15. Open, closed depend on $X$ and $𝒯$.

Example 16. $X=(0, 1)$ with usual metric, $\left[ \frac{1}{2}, 1 \right)$ is closed in $X$ as $X∖K$ is open but $K$ is not closed in $ℝ$.

Example 17. In the cofinite topology $K$ is closed ⇔ $K=X$ or $K$ is finite.

Proposition 18. Let $X$ be a topological space.

(C1) $∅, X$ are closed

(C2) $K_1, K_2$ closed ⇒ $K_1∪K_2$ closed

(C3) $K_i$ closed, $i∈ℐ$ ⇒ $\bigcap_{i∈ℐ}K_i$ is closed

Proof. (C1) $X=X∖∅$ and $∅$ is open.

(C2) $X∖(K_1∪K_2)=\underbrace{(X∖K_1)}_{\text{open}}∩\underbrace{(X∖K_2)}_{\text{open}}⇒$open by T2

(C3) $X∖\left( \bigcap_{i∈ℐ}K_i \right)=⋃_{i∈ℐ}(X∖K_i) ⇒$closed by T3□

Remark 19. We could define a topology giving the closed sets instead of open sets.

Definition 20. A sequence $(x_n)$ in a topological space $X$ converges to $x∈X$ if for any open $U \ni x$, $∃N$ such that $∀n≥N, x_n∈U$.

Remark 21. This is equivalent to definition for metric space $(X, d)$.

Example 22. $X=\{1, 2, 3 \}, 𝒯=\{∅, \{1 \}, \{1, 2 \}, \{1, 2,3 \}\}$

$x_n=1 ∀ n$. Then $x_n→1, x_n→2, x_n→3$. This does not happen if $X$ is Hausdorff.

Example 23. If $X$ is indiscrete space, any $(x_n)$ converges to $x∈X$.

Example 24. $X$ infinite set with cofinite topology. Let $(x_n)$ be a sequence such that all $x_n$ are distinct, then $x_n→x$ for all $x∈X$.

Corollary 25. $X$ indiscrete and $X_∞$ cofinite are not metrizable, because then limits would be unique.

Proposition. $X$ is a topological space. $F⊆X$ is closed then for any $(x_n)⊆F$ any limit of $(x_n)$ lies in $F$.

Proof. Let $a∈X∖ F = U$. $U$ is open.

Assume $x_n→a$. Then $∃N$ such that $x_n∈U = X∖F \quad ∀n≥N$

But $x_n∈F$, impossible◻

Remark. Converse not true (Exercise on PS1).

Definition. Let $(X, 𝒯_X), (Y, 𝒯_Y)$ be topological spaces and $f : X→Y$ be a function.

$f$ is continuous if $∀U∈𝒯_Y, f^{- 1} (U)∈𝒯_X$

Remark. If $(X, d)$ is a metric space, this is equivalent to the $(\varepsilon, \delta)$-definition of continuity.

Proposition. $f : X→Y$ is continuous iff for any $K⊆Y$ closed, $f^{- 1} (K)$ is closed.

Proof. ($⇒$) Say $K⊆Y$ closed. Then $Y∖K$ is open, so $f^{- 1} (Y∖K) = X∖f^{- 1} (K)$ is open$⇒ f^{- 1} (K)$ is closed.

($⇐$) Let $U⊆Y$ be open. Then $K = Y∖U$ is closed and $U = Y∖K$, so $f^{- 1} (U) = f^{- 1} (Y∖K) = X∖f^{- 1} (K), f^{- 1} (K) \text{ is open}⇒f^{- 1} (U)$ is open$⇒ f$ is continuous.◻

Proposition. If $X \xrightarrow{f} Y \xrightarrow{g} Z$, $X, Y, Z$ are topological spaces, $f, g$ continuous, then $g \circ f : X→Z$ is continuous.

Proof. $U⊆Z$ is open, then $(g \circ f)^{- 1} (U) = f^{- 1} (g^{- 1} (U))$ is open$⇒ g \circ f$ is continuous.◻

Proposition. If $f : X→Y$ is continuous map between topological spaces $X, Y$. If $x_n→x$ in $X$, then $f (x_n)→f (x)$.

Proof. Say $U$ is open such that $f (x)∈U$, $f^{- 1} (U)$ is open, $x∈f^{- 1} (U)$.

Since $x_n→x$, $∃N$ such that $∀n≥N, x_n∈f^{- 1} (U)⇒f (x_n)∈U$. So $f (x_n)→f (x)$.◻

Remark. Converse not true. 连续性的标准定义

Definition. A homeomorphism between topological spaces $X, Y$ is a bijection $f : X→Y$ such that $f, f^{- 1}$ are both continuous.

Remark. $f : X→Y$ a bijection between $(X, 𝒯_X), (Y, 𝒯_Y)$ is a homeomorphism iff $U∈𝒯_X ⇔ f (U)∈𝒯_Y$.

Write $X∼Y$ if $X, Y$ are homeomorphic.

Example.

$[0, 1]∼[0, 2]$ by $f (x) = 2 x$
$(0, 1)∼(a, b), a \neq b$
$(0, 1)∼(1, ∞)$ by $1 / x$
$(1, ∞)∼(0, ∞)$ by $x - 1$
$ℝ∼(0, ∞)$ by $e^x$
$f : [0, 1)→S^1 \quad f (x) = e^{2 nix}$ is continuous, 1-1, onto, but $f^{- 1}$ is not continuous, so $f$ is not a homeomorphism. In fact, $[0, 1), S^1$ are not homeomorphic.

Problem. Find $X, Y$ topological spaces such that $∃f : X→Y, g : Y→X$ continuous bijections but $X≁Y$. MSE

Definition. Let $X, Y$ be topological spaces and $A⊆X$. The closure of $A$ is the set

\[\overline A = \bigcap_{\substack{ F \text{ closed}\\ A⊆F }} F \quad \text{note } A⊆X, X \text{ closed, so }\overline A⊇A\]

Proposition 1 . $\overline A$ is the smallest closed set containing $A$, i.e.

(a) $A⊆\overline A$

(b) $∀F$ closed such that $A⊆F,\overline A⊆F$

Proof. By definition.◻

Proposition 2 . Let $A, B⊆X, X$ topological space.

(a) $A⊆B⇒\overline A⊆\overline B$

(b) $A$ is closed$⇔\overline A=A$

Proof.

(a) $A⊆B⊆\overline B,\overline B$ is closed. By Prop 1b, $\overline A⊆\overline B$.

(b) $A$ closed, $A⊆A⇒\overline A⊆A$ by Prop 1b. But $A⊆\overline A$, so $A =\overline A$.

Proposition 3 . $\overline A = \{ x∈X : ∀U \text{ open with } x∈U, U \cap A \neq ∅ \}$

Proof. Let $A_1 = \{ x∈X : ∀U \text{ open with } x∈U, U \cap A \neq ∅ \}$.

Clearly $A⊆A_1$.

To show $A_1$ is closed, it is enough to show that $X∖A_1$ is open.

Consider $x∈X∖A_1$. Then $∃U_x$ open, $x∈U_x, U_x \cap A = ∅$.

Moreover if $y∈U_x$ then $y∉A_1$ since $y∈X∖A_1$.

So $X∖A_1 = \bigcup_{x∉A_1} U_x$ is open so $A_1$ is closed.

$A⊆A_1$ is closed, so by Prop 1, $\overline A⊆A_1$.

Conversely, I claim $A_1⊆\overline A ⇔$If $x∉\overline A$ then $x∉A_1$.

$∃K \supseteq A$ closed such that $x∉K$.

Then $x∈X∖K, X∖K$ is open, $A⊆K$ and $(X∖K) \cap A = ∅$ so $x∉A_1$.◻

Example.

$\overline{(a, b)}$ in $ℝ$ is $[a, b]$
If $X = (a, b)$, then $\overline{(a, b)} = (a, b)$
$\overline{ℚ} = \overline{ℝ∖ℚ} =ℝ$ (in $ℝ$) by Prop 3
$X = \{ 1, 2, 3 \}, 𝒯= \{ ∅, \{ 1 \}, \{ 1, 2 \}, \{ 1, 2, 3 \} \}$
$\overline{\{ 1 \}} = \{ 1, 2, 3 \}$
($X$, discrete topology) $\overline{A} = A \quad ∀A⊆X$
($X$, indiscrete topology) $\overline{A} = X \quad ∀A \neq ∅$
($X$ infinite, cofinite topology) If $A$ finite, then $\overline A=A$; if $A$ infinite, then $\overline{A}=X$

Proposition. A map of topological spaces $f : X→Y$ is continuous iff $f (\overline{A})⊆\overline{f (A)} \quad ∀A⊆X$.

Proof. Assume $f$ continuous, $\overline{f (A)}$ is closed and $A⊆f^{- 1} (\overline{f (A)})$ where $f^{- 1} (\overline{f (A)})$ is closed.

By Prop 1, $\overline{A}⊆f^{- 1} (\overline{f (A)})⇒f (\overline{A})⊆\overline{f (A)}$.

Conversely, let $K⊆Y$ be closed. Let $A = f^{- 1} (K)$.

Then $f (A)⊆K$ where $K$ is closed, by Prop 1, $\overline{f (A)}⊆K$.

By assumption $f (\overline{A})⊆\overline{f (A)}⊆K⇒\overline{A}⊆f^{- 1} (K)⇒\overline{A}⊆A$.

Also $A⊆\overline{A}$, so $A = \overline{A}$.◻

Proposition 1. $x∈\overline{A}⇒∀U$ open, $x∈U$: $U∩A≠∅$.

Definition. A point $x∈X$ is an accumulation point (or limit point) if $∀U$ open, $x∈U$, $(U∖\{ x \})∩A≠∅$. Otherwise if $x∈\overline{A}$ is not an accumulation point then $x$ is an isolated point of $A$.

Notation. $A'$ is the set of accumulation points of $A$. By Proposition 1, $\overline{A}=A∪A'$.

Example. In ℝ standard topology,

$A=\left\{ \frac{1}{n}:n∈ℕ \right\}, A'=\{ 0 \}$
$ℚ' =ℝ$
$A=(0, 1)∪\{ 3 \}, A'=[0, 1]$, 3 is isolated point

Proposition. Let $(X, d)$ be a metric space and let $A⊆X$

$\overline{A}$ is the set of limit points of convergent sequences $(a_n)$ in $A$.
$A'$ is the set of limits of convergent sequences $(a_n)$ with pairwise distinct terms ($a_n≠a_m$ if $n≠m$)

Proof.

$x∈\overline{A}$, by Prop 1, take $a_1∈X$, take $a_n∈B (x, r_n)$, where $r_n=\min \left( \frac{1}{n}, d (a_{n - 1}, x) \right)$.
If $a_n→x$ and $a_n$ are pairwise distinct, only one of them can be $x$$⇒U∖\{ x \}∩A≠∅⇒x∈A'$.
Pick $a_n∈B (x, r_n)$, where $r_n=\min \left( \frac{1}{n}, d (a_{n - 1}, x) \right)$.

◻

Definition. Let $X$ be a topological space, $A⊆X$. The interior of $A$ is the set $\mathring{A}=\bigcup_{\substack{V \text{ open}\\ V⊆A}} V$

Example.

$[0, 1]^{\circ}=(0, 1)$ in ℝ.
$\mathring{A} =\mathring{ℝ∖ℚ}=∅$
$A =ℝ×\{ 0 \}⊆ℝ×ℝ,\mathring{A}=∅$
$X$ infinite with co-finite topology, $A⊆X$ such that $X∖A$ is infinite then $\mathring{A}=∅$.

Remark. $\mathring{A}$ is the largest open subset of $X$ contained in $A$.

$\mathring{A}$ is open in $X$, $\mathring{A}⊆A$.
$∀U$ open, $U⊆A$, $U⊆\mathring{A}$.

Proof. By definition◻

Proposition. Let $A, B$ subsets of topological space $X$. Then

$A⊆B⇒\mathring{A}⊆\mathring{B}$
$A$ is open$\Leftrightarrow A =\mathring{A}$
$\mathring{\mathring{A}} =\mathring{A}$

Proof.

$\mathring{A}⊆A⊆B$, so $\mathring{A}$ is open, contained in $B$, so $\mathring{A}⊆\mathring{B}$.
$A$ is open$⇒A⊆\mathring{A}$. But $\mathring{A}⊆A⇒A =\mathring{A}$
$A =\mathring{A}$ and $\mathring{A}$ is open$⇒A$ is open
$\mathring{A}$ is open, so by b) $\mathring{\mathring{A}} =\mathring{A}$

◻

Proposition 2. Let $A⊆X$ and $X$ be a topological space. Then $\overline{X∖A}=X∖\mathring{A}$.

Proof. $X∖\mathring{A}=X∖\left( \bigcup_{\substack{V \text{ open}\\ V⊆A}} V \right)=\bigcap_{\substack{V \text{ open}\\ V⊆A}} (X∖V)$ and $X∖V$ is closed.

$V⊆A⇒X∖V⊇X∖A$.

$X∖\mathring{A}=\bigcap_{\substack{F \text{ closed}\\ F \supseteq X∖A}}F= \overline{X∖A}$◻

Proposition 3. Say $B⊆X$, $X$ is a topological space. Then $\mathring{X∖B}=X∖\overline{B}$.

Proof. Set $B=X∖A$, then $A=X∖B$, $\mathring{A} =\mathring{X∖B}$ and $X∖\mathring{A}=\overline{X∖A}=\overline{B}⇒X∖\overline{B} =\mathring{A} =\mathring{X∖B}$◻

Definition. If $A⊆X$ topological space, the set $∂A=\overline{A}∖\mathring{A}$ is the boundary of $A$.

Example. In ℝ, $A=(0, 1)$, $\overline{A}=[0, 1]$, $\mathring{A}=(0, 1)$, $∂A=\{ 0, 1 \}$. $∂B (0, 1)=S^1$.

Proposition. The boundary of $A$ is equal to $\overline{A}∩\overline{X∖A}$

Proof. $∂A=\overline{A}∖\mathring{A}=\overline{A}∩(X∖\mathring{A}) \mathop{=}\limits_{\text{Prop. 2}} \overline{A}∩\overline{X∖A}$◻

Corollary. $∂A=∂(X∖A)$

Proof. $∂A=\overline{A}∩\overline{X∖A}$ is the same as $∂(X∖A)=\overline{X∖A}∩\overline{X∖(X∖A)}=\overline{X∖A}∩\overline{A}$◻

Example. $\overline{∂ℚ}=\overline{ℚ}∩\overline{ℝ∖ℚ} =ℝ$

Exercise. Find 3 disjoint sets $A_1, A_2, A_3$ in ℝ² such that $∂A_1=∂A_2=∂A_3$

Lakes of Wada

Separation Axioms

Definition. A topological space $X$ satisfies the first separation axiom if for any $a, b∈X$, $∃U$ open such that $a∈U, b∉U$.

Example. $X$ infinite with co-finite topology satisfies the separation axiom.

Let $a, b∈X, a≠b$. Then $X∖\{ b \}∋a$ is open.

Proposition. A topological space $X$ satisfies the first separation axiom iff singletons are closed.

Proof. (⇐) Say singletons are closed. Take $a≠b$. $U=X∖\{ b \}$ is open since $\{ b \}$ is closed, $a∈U, b∉U$.

(⇒) Let $b∈X∖\{ a \}$, $∃V$ open, $b∈V⊆X∖\{ a \}$. So $X∖\{ a \}$ is open, so $\{ a \}$ is closed.◻

Definition. A topological space $X$ is said to be Hausdorff(or satisfy second separation axiom) if given any $a≠b$ in $X$ there are disjoint open sets $U, V$ such that $a∈U$ and $b∈V$.

Example.

Any metric space is Hausdorff.
$X$ infinite with cofinite topology is not Hausdorff.

Proposition. Let $X, Y$ be topological spaces. If $f:X→Y$ is injective, continuous and $Y$ satisfies the first (or second) separation axiom, then $X$ satisfies the first (or second) separation axiom.

Proof.

$a≠b∈X$, then $f (a)≠f (b)∈Y$, $∃U$ open in $Y$ such that $f (a)∈U$ and $f (b)∉U$.
By continuity, $f^{- 1} (U)$ is open, $a∈f^{- 1} (U)$ and $b ∉ f^{- 1} (U)$.
$a≠b∈X$, then $f (a)≠f (b)∈Y$, $∃U$ open in $Y$ such that $f (a)∈U, f (b)∈V, U∩V =∅$. So $f^{- 1} (U)∩f^{- 1} (V) =∅$.
By continuity, $f^{- 1} (U), f^{- 1} (V)$ are open, $a∈f^{- 1} (U), b∈f^{- 1} (V)$.◻

Proposition. $f:X→Y$ is continuous injection, $Y$ satisfies first (second) separation axiom, then $X$ satisfies first (second) separation axiom. (proved at the end of lecture 3)

Corollary. If $X, Y$ are homeomorphic topological spaces, then $X$ satisfies first (second) separation axiom iff $Y$ satisfies first (second) separation axiom.

Proposition. In a Hausdorff space, a sequence converges to at most one limit.

Proof. Say $x_n→a, b≠a$. By Hausdorff, ∃disjoint open sets $U∋a, V∋b$, then $∃N∀n⩾N:x_n∈U⇒x_n∉V⇒x_n↛b$.◻

Example. Let $X$ be infinite with cofinite topological space, $a∈X∖\{b\}$. Then $X$ satisfies first separation axiom, but limit not unique:

Let $(x_n)$ be a sequence such that $x_i≠x_j$ if $i≠j$. Then $∀b∈X, x_n→b$.

Subspaces

Definition. Let $(X,𝒯)$ be a topological space and let $A$ be non-empty subset of $X$. The subspace (or induced) topology on $A$ is $𝒯_A = \{A∩U:U∈𝒯\}$.

Check that 𝒯 satisfies T1,T2,T3.

T1) $∅∈𝒯_A, A = X∩A∈𝒯_A$

T2) $A∩V∈𝒯_A, A∩V∈𝒯_A$. Then $(A∩U)∩(A∩V) = A∩(U∩V)∈𝒯_A$

T3) $\bigcup_{i∈I} (A∩U_i) = A∩\bigcup_{i∈I} U_i∈𝒯_A$

Lemma. Let $(X, 𝒯)$ be a topological space, $A⊆X$. Then $i:A→X, i (a) = a$ is continuous for $(A, 𝒯_A)$.

Proof. If $U⊆X$ is open, then $i^{-1}(U) = A∩U∈𝒯_A$.◻

Lemma. Let $(X, 𝒯)$ be a topological space, $A⊆X$. Let $f:X→Y$ be continuous. Then $f |_A$ is continuous for $(A, 𝒯_A)$.

Proof. We have composition of maps: $A \xrightarrow{i} X \xrightarrow{f} Y$

$f|_A = f∘i⇒f|_A$ is continuous.◻

Proposition 1. Let $(X, 𝒯)$ be a topological space, $A⊆X$. $W⊆A$ is closed in $𝒯_A$ iff $∃F$ closed in $(X, 𝒯)$ such that $W = F∩A$.

Proof. $W$ is closed in 𝒯$⇔A∖W = A∩U, U∈𝒯$.

$⇔W=A∖(A∩U) = (A∖U)∩A = (X∖U)∩A, W = F∩A, F = X∖U$ closed.◻

Example. $ℚ⊆ℝ$

$\mathring{ℚ} =ℚ, \overline{ℚ} =ℚ$ in 𝒯_ℚ

$\mathring{ℚ} = ∅, \overline{ℚ} =ℝ$ in 𝒯_ℝ

Proposition 2. Let $(A, 𝒯_A)$ be a subspace of $(X, 𝒯)$ and let $B⊆A$

$\overline{B}^A = \overline{B}^X∩A$
$\mathring{B}^A⊇\mathring{B}^X$

Proof.

(⇒) $\overline{B}^X = F$ is a closed subset of $X$, $B⊆\overline{B}^X$, by Prop.1 $F∩A$ is a closed subset of $A$, and $F∩A⊇B$.
So $\overline{B}^X∩A⊇\overline{B}^A$.
(⇐) Since $\overline{B}^A$ is closed, by Prop.1 $\overline{B}^A = F∩A$ for some $F$ closed in $X$.
$\overline{B}^A⊇B⇒F⊇B⇒F⊇\overline{B}^X⇒\overline{B}^A = F∩A⊇\overline{B}^X∩A$.◻
Since $\mathring{B}^X$ is open in $X$ and contained in $B⊆A$, it is an open set in $A$. Therefore $\mathring{B}^A⊇\mathring{B}^X$.

Example. $A = \{0, 1\}⊆[0, 1] = X$

$B = \{0\}$ then $\{0\} = \mathring{B}^A≠\mathring{B}^X = ∅$

Proposition. If $A⊆X$ is open, then $∀B⊆A, \mathring{B}^A = \mathring{B}^X$

Proof. $\mathring{B}^A=U∩A$ open in $X$, also $\mathring{B}^A⊆A⇒\mathring{B}^A⊆\mathring{B}^X$◻

Remark. on subspace topology

If $A⊆B⊆X$, $(X, 𝒯)$ is topological space, then the topology induced on $A$ by 𝒯 coincides with the topology induced on $A$ by $𝒯_B$.
Let $U∈𝒯$.
$U∩A$ is an open set in the topology induced on $A$ by 𝒯
$(U∩B)∩A$ is an open set in the topology induced on $A$ by $𝒯_B$
Note that $U∩A=(U∩B)∩A$.
If $(X, d)$ is a metric space, $A⊆X$, then $(A, d)$ is a metric space.
Then $𝒯_A$ is the topology induced on $A$ by $d$.
For $d$: $U$ is open$⇔U$ is union of open balls for $d$.
$B^A (a, r) = B^X (a, r)∩A$
$B^A (a, r)$ open in $d$
$B^X (a, r)∩A$ open in $𝒯_A$
$A$ is open$⇒ 𝒯_A⊆𝒯$.
$A$ is closed$⇒$if $F⊆A$ is closed in $𝒯_A$, then $F$ is closed in 𝒯.

Basis for a topology

Definition. Given a topological space $(X, 𝒯)$, a collection of subsets ℬ of $X$ is called a basis for 𝒯 if

$ℬ⊆𝒯$
Every $U∈𝒯$ can be written as union of elements of ℬ

Example.

$(X, d)$ metric space
$ℬ= \{B (x, r):x∈X, r∈ℝ_+\}$ is a basis.
for $ℝ$, $ℬ= \{(a, b):a, b∈ℝ\}$ is a basis
for $ℝ^2$, $ℬ= \{B (a, r):a∈ℚ×ℚ, r∈ℚ^+\}$ is a countable basis

Criterion. Let $(X, 𝒯)$ be a topological space. ℬ is a basis. $U$ is open$⇔∀x∈U ∃B∈ℬ$ such that $x∈B, B⊆U$.

Proof. ($⇒$) $U$ open$⇒U=\bigcup_{i∈I} B_i, B_i∈ℬ ⇒$if $x∈U$ then $x∈B_i$ for some $i$.

($⇐$) $∃B_x∈ℬ, x∈B_x$, then $U = \bigcup_{x∈U} B_x⇒U$ is open.◻

Proposition. Let $X, Y$ be topological spaces, ℬ is a basis for topology of $Y$. Then $f:X→Y$ is continuous iff $∀B∈ℬ: f^{-1}(B)$ is open.

Proof. ($⇒$) If $f$ is continuous then $f^{-1}(B)$ is open $∀B∈ℬ$.

($⇐$) $U = \bigcup_{i∈I} B_i, f^{-1}(U) = \bigcup_{i∈I} f^{-1}(B_i)$ is open.◻

Proposition. Let $X$ be a set, ℬ a family of subsets such that

(B1) $X = \bigcup_{B∈ℬ} B$

(B2) $∀B_1, B_2∈ℬ, B_1∩B_2$ is a union of elements of ℬ

Then the family $𝒯_ℬ$ of unions of sets in ℬ defines a topology on $X$.

Proof. (T1) $∅, X∈𝒯_ℬ$

(T2) $\bigcup_{i∈I} B_i∩\bigcup_{j∈J} B_j = \bigcup_{\substack{i∈I\\ j∈J}} (B_i∩B_j)∈𝒯_ℬ$

(T3) $\bigcup_{j∈J} \left( \bigcup_{i∈I} B_{j,i} \right)∈𝒯_ℬ$◻

Proposition. Let $(X, 𝒯_X), (Y, 𝒯_Y)$ be topological spaces, then the sets $ℬ_{X×Y}=\{U×V:U⊆X, V⊆Y \text{ open}\}$ satisfy conditions (B1),(B2). So it is a basis for a topology called product topology.

Remark. $\{U×V:U, V \text{ open}\}$ is not a topology: doesn't satisfy (T3): the union of 2 rectangles is not a rectangle.

Proof. (B1) $X∈𝒯_X, Y∈𝒯_Y⇒X×Y∈ℬ_{X×Y}$

(B2) $(U_1×V_1)∩(U_2×V_2)∋(x, y)⇔x∈U_1∩U_2, y∈V_1∩V_2$

So $(U_1×V_1)∩(U_2×V_2)=(U_1∩U_2)×(V_1∩V_2)∈ℬ_{X×Y}$◻

Definition. The space $(X×Y, 𝒯_{X×Y})$ is called the topological product of $(X, 𝒯_X), (Y, 𝒯_Y)$.

Proposition. Let $(X×Y, 𝒯_{X×Y})$ be the topological product of $(X, 𝒯_X), (Y, 𝒯_Y)$. If $A$ is closed in $𝒯_X$, $B$ is closed in $𝒯_Y$, then $A×B$ is closed in $𝒯_{X×Y}$.

Proof. $(X×Y)∖(A×B)∋(x, y)⇔x∉A \text{ or } y∉B$

So $(X×Y)∖(A×B)=(X∖A)×Y∪X×(Y∖B)∈𝒯_{X×Y}$

So $A×B$ is closed.◻

Remark. Similarly define product topology for $X_1×X_2×⋯×X_n$ using $U_1×U_2×⋯×U_n$ as basis for $U_i$ open.

Proposition. The product topology on $ℝ^n$ is equal to the standard topology on $ℝ^n$.

Proof. Recall $d_1, d_2, d_∞$ give the same topology on $ℝ^n$. Here it is more convenient to use $d_∞$.

\[d_∞ ((x_1,…, x_n), (y_1,…, y_n))=\max_{1⩽i⩽n}{|x_i-y_i|}\]

Say $W$ is an open set in product topology.

Let $x=(x_1,…, x_n)∈W$, then $∃U_1,…, U_n$ open such that $x∈U_1×⋯×U_n⇒∀i:x_i∈U_i⇒∃ε_i:(x_i - ε_i, x_i + ε_i)∈U_i$

Let $ε=\min_{1⩽i⩽n} ε_i$, then $B (x, ε)⊆W⇒W$ is a union of $d_∞$-balls, so $W$ is open in standard topology.

In the other direction, say $W$ is open in standard topology, so $W$ is a union of open balls $B ((x_1,…, x_n), ε)=(x_1 - ε, x_1 + ε)×⋯×(x_n - ε, x_n + ε)∈ℬ_{X×Y}$, so $W∈𝒯_{X×Y}$.◻

Proposition. If topological spaces $X, Y$ satisfy 1st(or 2nd) separation axiom, so does $X×Y$.

Proof. (1st sep axiom) Let $(x_1, y_1), (x_2, y_2)∈X×Y$. Say $x_1≠x_2$, $∃U$ open in $X, x_1∈U, x_2∉U$. Then $(x_1, y_1)∈U×Y, (x_2, y_2)∉U×Y$. Similarly if $y_1≠y_2$.

(2nd sep axiom) Let $(x_1, y_1), (x_2, y_2)∈X×Y$. Say $y_1≠y_2$, $∃V_1, V_2$ open in $Y$, $V_1∩V_2 =∅$, such that $y_1∈V_1, y_2∈V_2$.

$(x_1, y_1)∈X×V_1, (x_2, y_2)∈X×V_2$ and $(X×V_1)∩(X×V_2) =∅$. Similarly if $x_1≠x_2$.◻

Proposition 1. a) Let $X, Y$ topological spaces and let $(X×Y, 𝒯_{X×Y})$ be their topological product. Then $p_X:X×Y→X, p_X (x, y)=x ; p_Y:X×Y→Y, p_Y (x, y)=y$ are continuous.

b) The product topology is the coarsest topology on $X×Y$ such that $p_X, p_Y$ are continuous.

Proof. a) $U⊆X$ open, $p_X^{-1}(U)=U×Y∈𝒯_{X×Y}$, so $p_X$ is continuous. Similarly $p_y$ is continuous.

b) Let $𝒯$ be a topology on $X×Y$ such that $p_X, p_Y$ are continuous. Then $p_X^{-1}(U)∈𝒯⇒U×Y∈𝒯, p_Y^{-1}(V)∈𝒯⇒X×V∈𝒯$, so $(U×Y)∩(X×V)=U×V∈𝒯$.

$U⊆X$ open, $V⊆Y$ open, so $𝒯⊇ℬ_{X×Y}⇒𝒯⊇𝒯_{X×Y}$◻

Proposition 2. Let $X, Y$ be topological spaces.

Let $f:Z→X, g:Z→Y$, then $F=(f, g):Z→X×Y$ is continuous where $X×Y$ is given product topology iff $f, g$ are continuous.
The product topology is the only topology for which (1) holds for all $Z, f, g$.

Proof.

Assume $(f, g)$ is continuous. $U$ is open in $X$, so $U×Y$ is open, $(f, g)^{-1}(U×Y)=f^{-1}(U)$ is open, so $f$ is continuous. Similar for $g$.
In the other direction, assume $f, g$ are continuous, $∀U×V∈ℬ_{X×Y},z∈(f, g)^{-1}(U×V)⇔f (z)∈U, g (z)∈V$
So $(f, g)^{-1}(U×V)=f^{-1}(U)∩g^{-1}(V)$ is open. So $(f, g)$ is continuous.
Let $𝒯$ be a topology on $X×Y$ for which (1) holds.
Take $Z=(X×Y, 𝒯)$. $f:X×Y→X, f=p_X ; g:X×Y→Y, g=p_Y$.
Consider $(f, g):(X×Y, 𝒯) \xrightarrow[\text{id}]{} (X×Y, 𝒯)$
Clearly id is continuous, by (1), $p_X, p_Y$ are continuous, by Prop 1, $𝒯_{X×Y}⊆𝒯$.
Take $Z=(X×Y, 𝒯_{X×Y})$. Then $p_X, p_Y$ are continuous, $p_X:Z→X, p_Y:Z→Y$.
By property (1) of $𝒯$, $(p_X, p_Y):(X×Y, 𝒯_{X×Y}) \xrightarrow[\text{id}]{} (X×Y, 𝒯)$ is continuous. $(p_X, p_Y)^{-1}(U) =\operatorname{id}^{-1}(U)=U$ open in $𝒯_{X×Y}$. So $𝒯⊆𝒯_{X×Y}$. So $𝒯=𝒯_{X×Y}$.

◻

Definition. Let $X, Y$ be sets, $X⊔Y=(X×\{0\})∪(Y×\{1\})$.

Definition. Let $X, Y$ be topological spaces. We equip $X⊔Y$ with topology:

$U$ open in $X⊔Y$ iff $U∩(X×\{0\})$ is open in $X$ and $U∩(Y×\{1\})$ is open in $Y$.

Connected spaces

Definition. A topological space $X$ is disconnected if $∃U, V$ open, disjoint, non-empty such that $X=U∪V$. If $X$ is not disconnected, it is called connected.

Proposition. Let $X$ be a topological space. The following are equivalent:

The only subsets of $X$ that are both open and closed are $X, ∅$.
$X$ is connected.
Any $f:X→\{0, 1\}$ continuous is constant.

Proof. Seen in Metric Spaces.◻

Definition. A non-empty subset $A$ of a topological space is connected if $A$ with the subspace topology is connected.

Remark. ∅ is connected by convention.

Remark. Equivalent definition: $A$ is connected if $∀U, V$ open in $X$ such that $A⊂U∪V$ and $A∩U∩V=∅$, either $U∩A=∅$ or $V∩A=∅$.

Example. ℝ with cofinite topology.

$A=\{1, 2\}$ is not connected. $U=ℝ∖\{2\}, V=ℝ∖\{1\}, U∩A=\{1\}$ and $V∩A=\{2\}$ are open in $A$.

On the other hand, for every $U, V$ open in (ℝ, cofinite) such that $1∈U, 2∈V, U∩V≠∅$.

Proposition. If $f:X→Y$ is continuous, $A⊂X$ is connected, then $f (A)$ is connected.

Proof. Seen in Metric Spaces◻

Example. $[0, 1], S^1$ are both connected but not homeomorphic. Say $f:[0, 1]→S^1$ is a homeomorphism, then $f|_{[0, 1]-\left\{\frac12\right\}}:\left[0, \frac12\right)∪\left(\frac12, 1\right]→S^1 ∖\left\{f \left(\frac12\right)\right\}$ is homeomorphism, contradiction since $S^1∖\left\{f \left(\frac12\right)\right\}$ is connected but $\left[0, \frac12\right)∪\left(\frac12, 1\right]$ is not.

Proposition. Suppose $\{A_i:i∈I\}$ is a family of connected subsets of a topological space $X$ such that $A_i∩A_j≠∅ ∀i, j∈I$. Then $⋃_{i∈I}A_i$ is connected.

Proof. Let $f:⋃_{i∈I}A_i→\{0, 1\}$ be continuous. Since $A_i$ is connected, $f (A_i)=c_i∈\{0, 1\}$, since $A_i∩A_j≠∅$, let $x∈A_i∩A_j$, then $c_i=f (x)=c_j$, so $c_i=c_j ∀i, j∈I$, so $f$ is constant, so $⋃_{i∈I}A_i$ is connected.◻

Corollary. If $\{C_i:i∈I\}$ and $B$ are connected subsets of a topological space $X$ and $C_i∩B≠∅ ∀i$, then $\left(⋃_{i∈I}C_i\right) ∪B$ is connected.

Proof. Take $A_i=C_i∪B, A_i∩A_j \supset B$, so $A_i∩A_j≠∅$.◻

Example. $A⊂ℝ^2$

$A=\{(x, y):x∈ℝ, y∈\mathbb{Q}\}∪\{(0, x):x∈ℝ\}$

By the corollary, $A$ is connected.

Theorem. If and only if $X, Y$ are connected, the topological product $X×Y$ is connected.

Proof. ($⇒$) $p_X:X×Y→X$ is continuous, so $p_X (X×Y)=X$ is connected and similar for $Y$.

($⇐$) Note: For any $y_0∈Y$ fixed, the map $i:X→X×Y$ given by $i (x)=(x, y_0)$ is continuous.

Proof: If $U×V$ is open with $y_0∉V$, $i^{-1}(U×V)=∅$
If $U×V$ is open with $y_0∈V$, $i^{-1}(U×V)=U$ open

So $X×\{y\}$ is connected $∀y∈Y$.

$B=\{x_0\}×Y$ is connected.

$B∩(X×\{y\})=\{(x_0, y)\}≠∅$

$X×Y=B∪⋃_{y∈Y}(X×\{y\})$ is connected by the corollary.◻

Remark. $X_1×⋯×X_n$ is connected iff $X_i$ connected $∀i$.

Theorem. Suppose $A$ is a connected subset of a topological space $X$ and $A⊂B⊂\overline{A}$. Then $B$ is connected.

Proof. Let $U, V$ be open in $X$ such that $B⊂U∪V$ and $B∩U∩V=∅$. It is enough to show that $B∩U$ or $B∩V$ is empty.

Since $B∩U∩V=∅$ we have $A∩U∩V=∅$, but $A$ is connected, so either $A∩U=∅$ or $A∩V=∅$, say $A∩V=∅$. Then $A⊂X∖V$ closed$⇒ \overline{A}⊂X∖V$.

$B⊂\overline{A}⇒B⊂X∖V⇒B∩V=∅$.◻

Definition. A path $P$ connecting 2 points $x,y$ in $A$ is a continuous map $p:[0,1]→X$ such that $p(0)=x,p(1)=y$.

Remark. We know $[0,1]$ is connected, so $p([0,1])$ is connected.

Definition. A topological space $X$ is path connected if $∀a, b∈X$ there is a path connecting $a, b$.

$A⊂X$ is path connected if $∀a, b∈A, ∃p:[0, 1]→A$ continuous such that $p (0)=a, p (1)=b$.

Proposition. If $X$ is path connected, then $X$ is connected.

Proof. Fix $x_0∈X$. For any $y∈X$, take $p_y:[0, 1]→X$ such that $p (0)=x_0, p (1)=y$. $P_y=p_y ([0, 1])$.

$X=⋃_{y∈X}P_y, P_y∩P_z⊃\{x_0\}$, so $P_y∩P_z≠∅$, so by proposition $X$ is connected.◻

Example. The image of $⟨(1+1/t)\cos t,(1+1/t)\sin t⟩$ along with the unit circle is connected in $ℝ^2$ but not path connected. MSE

Proposition. Let $f:X→Y$ be continuous map of topological spaces, $A⊂X$ path connected. Then $f (A)$ is path connected.

Proof. If the path $p:[0, 1]→X$ joins $a, b$, the path $f∘p$ joins $f (a), f (b)$.◻

Compact spaces

Definition. A family $\{U_i:i∈I\}=𝒰$ of subsets is called a cover of $X$ if $X=⋃_{i∈I}U_i$. If each $U_i$ is open in $X$, $𝒰$ is called an open cover of $X$. A subcover of a cover $\{U_i:i∈I\}$ is a subfamily $\{U_i:i∈J\}$ for some $J⊂I$ such that $X=⋃_{j∈J}U_j$.

We say that this subcover is finite(countable) if $J$ is finite(countable).

Definition. A topological space $X$ is compact if every open cover of $X$ has a finite subcover.

A subset $A$ of $X$ is compact if $A$ is compact if $A$ is compact with the subspace topology.

$\{U_i:i∈I\}$ is a cover of $A⊂X$ if $A⊂⋃_{i∈I}U_i$ (Similarly for open cover, subcover)

Remark. $A$ is compact if every open cover of $A$ has a finite subcover.

Proposition. Let $X$ be a topological space, the following are equivalent:

$X$ is compact
If $\{V_i:i∈I\}$ is a family of closed subsets of $X$ s.t. $\bigcap_{i∈I} V_i=∅$ then $∃$ finite $J⊆I$ s.t. $\bigcap_{j∈J} V_j=∅$

Proof. (1)$⇒$(2) Let $V_i, i∈I$ be s.t. $\bigcap_{i∈I} V_i=∅$.

$\bigcup_{i∈I} (X∖V_i)=X∖\bigcap_{i∈I} V_i=X$

So $\{X∖V_i:i∈I\}$ is open cover of $X$.

$X$ compact$⇒ X=(X∖V_{i_1})∪⋯∪(X∖V_{i_n})=X∖\bigcap_{j=1}^n V_{i_j}⇒ \bigcap_{j=1}^n V_{i_j}=∅$ i.e. (2) holds.

(2)$⇒$(1) Let $\{U_i:i∈I\}$ be open cover of $X$.

$\bigcup_{i∈I} U_i=X⇒ \bigcap_{i∈I} X∖U_i=∅, X∖U_i$ closed.

By (2) $∃U_{i_1},…, U_{i_n}$ s.t. $\bigcap_{j=1}^n X∖U_{i_j}=∅$

$X=\bigcup_{j=1}^n U_{i_j}$ finite cover$⇒ X$ compact. ◻

Definition. A subset $A$ of a topological space is compact if it is a compact topological space when endowed with the subspace topology.

Definition. $\{U_i:i∈I\}$ is a cover of $A$ if $A⊆\bigcup_{i∈I} U_i$

Equivalent definition of compact subset:

$A$ is compact$\Leftrightarrow$For every family $\{U_i:i∈I\}$ which is a cover of $A$ with $U_i$ open in $X$, $∃U_{i_1},…, U_{i_n}, i_j∈I$ s.t. $A⊆\bigcup_{j=1}^n U_{i_j}$

Example.

Any finite topological space is compact. A discrete space is compact if and only if it is finite.
The indiscrete space is compact
If a metric space $X$ is compact then $X$ is bounded
Proof. $X⊆\bigcup_{n∈ℕ} B (x_0, n)⇒ X⊆B (x_0, N)$ for some $N⇒ X$ is bounded. ◻
Let $X$ be any space with cofinite topology. Then any $A⊆X$ is compact.
Proof. Say $A⊆\bigcup_{i∈I} U_i$ where $U_i$ are open
If $i_0∈I$, $A∖U_{i_0}=\{a_1,…, a_k\}, a_j∈U_{i_j}, i_j∈I$. So $A⊆U_{i_0}∪U_{i_1}∪⋯∪U_{i_k}$. ◻
(Heine-Borel) $[a, b]∈\mathbb{R}$ is compact
Proof. see Metric Space course. ◻
Remark. $(a, b)$ is not compact: $(a, b)=\bigcup_{n∈ℕ} \left( a + \frac{1}{n}, b - \frac{1}{n} \right)$, ∄finite subcover

Proposition. Any closed subset $A$ of a compact space $X$ is compact.

Proof. Say $A⊆\bigcup_{i∈I} U_i, U_i$ open, then $X⊆(X∖A)∪\bigcup_{i∈I} U_i, X$ compact$⇒ X⊆(X∖A)∪(U_{i_1}∪⋯∪U_{i_n})$ for some $i_j∈I⇒ A⊆U_{i_1}∪⋯∪U_{i_n}⇒ A$ is compact. ◻

Remark. The converse is not true in general for topological spaces, so compact subsets might not be closed, e.g. $X$ with indiscrete topology. Then $\{x\}⊆X$ is compact but $\{x\}$ is not closed.

Proposition. Let $X$ be Hausdorff and $K⊆X$ be compact. If $x∈X∖K$, there are open disjoint sets $U, V$ such that $K⊆U, x∈V$.

Proof. For each $y∈K$ pick $U_y, V_y$ open disjoint such that $y∈U_y, x∈V_y, K⊆\bigcup_{y∈K} U_y⇒ K⊆U_{y_1}∪⋯∪U_{y_n}$. Let $V=\bigcap_{i=1}^n V_{y_i}$.
$∀i:V_y∩U_{y_i}=∅, V⊆V_{y_i}⇒V \cap U_{y_i}=∅⇒V∩\underbrace{(U_{y_1}∪⋯∪U_{y_n})}_U=∅$. Also $K⊆U$. ◻

Corollary. Any compact subset $K$ of a Hausdorff space $X$ is closed in $X$.

Proof. We show $X∖K$ is open

By Prop. $∀x∈X∖K∃U_x \ni x$ open such that $U_x \cap K=∅$.

So $X∖K$ is union of $U_x$(open)

So $X∖K$ is open$⇒ K$ is closed ◻

Proposition. If $f:X→Y$ is continuous map of topological spaces and $A⊆X$ is compact, then $f (A)$ is compact.

Proof. Say $f (A)⊆\bigcup_{i∈I} U_i, U_i$ open, then $A⊆\bigcup_{i∈I} f^{- 1} (U_i)$, $f^{- 1} (U_i)$ open since $f$ continuous.

Since $A$ compact, $A⊆f^{- 1} (U_{i_1})∪⋯∪f^{- 1} (U_{i_k})$ for some $i_j∈I⇒ f (A)⊆U_{i_1}∪⋯∪U_{i_n}⇒ f (A)$ compact. ◻

Remark. If $K⊆A⊆X$, $X$ topological space, $K$ compact in $A$ (with subspace topology) then $K$ is compact in $X$.

Proof. $i:A→X, i (a)=a$ is continuous. $K⊆A$ is compact$⇒ i (K)=K$ is compact in $X$. ◻

Remark. Not true for closed e.g. $A=(0, 2)⊆\mathbb{R}, K=(0, 1], K$ is closed in $A$ but not in $\mathbb{R}$. compactness is an inherent notion

The converse implication holds both for ‘closed’ and for ‘compact’, i.e. if $K⊆A⊆X$ and $K$ is closed (compact) in $X$, then $K$ is closed (compact) in $A$.

Proposition. Let $X$ be a topological space.

If $K_1,…, K_n$ are compact subsets of $X$ then $K_1∪⋯∪K_n$ is compact.
If $\{K_i, i∈I\}$ are compact and $X$ is Hausdorff then $\bigcap_{i∈I} K_i$ is compact.

Proof.

Say $\{U_i:i∈I\}$ is a cover of $K_1∪⋯∪K_n$. $∃J_1,…, J_n⊆I$ finite such that $K_r⊆\bigcup_{j∈J_r} U_j⇒ K_1∪⋯∪K_n⊆\bigcup_{r=1}^n\bigcup_{j∈J_r} U_j$
Since $X$ is Hausdorff, compact$⇒$closed, so $K_i$ closed$∀i⇒ \bigcap_{i∈I} K_i$ is closed.
Pick $K_{i_0}, i_0∈I, \bigcap_{i∈I} K_i⊆K_{i_0}$ (closed)
$K_{i_0}$ compact$⇒ \bigcap_{i∈I} K_i$ is compact in $K_{i_0}⇒ \bigcap_{i∈I} K_i$ is compact in $X$.

◻

Remark. (1) not true for infinite union e.g. $(0, 1)=\bigcup_{n∈ℕ} \left[ \frac{1}{n}, 1 - \frac{1}{n} \right]$

(2) not true if $X$ is not Hausdorff (Ex.5 sheet3)

Theorem. The product $X×Y$ is compact iff $X, Y$ are compact.

Proof. ($⇒$) easy $P_X (X×Y)=X⇒ X$ compact. Same for $Y$.

($⇐$) $X×Y⊆\bigcup_{i∈I} W_i$. Since the rectangular open sets compose a basis for the product topology, replace $W_i$ by sets $U×V$, $U$ open in $X$, $V$ open in $Y$.\[X×Y⊆\bigcup_{j∈J} U_j×V_j\]Therefore it is enough if we prove that $\{U_j×V_j:j∈J\}$ has a finite subcover.

For every $y∈Y$, the compact set $X×\{y\}$ is covered by $\{U_j×V_j:j∈J\}$. Therefore $∃$ finite subset $F_y⊆J$ such that\[X×\{y\}⊆\bigcup_{j∈F_y} U_j×V_j\]The set $V_y=\bigcap_{j∈F_y} V_j$ is an open set containing $y$. The family $\{V_y:y∈Y\}$ is an open cover of $Y$, which is compact.

It follows that there exist $y_1,…, y_m$ such that $Y=V_{y_1}∪⋯∪V_{y_m}$.

We state that $X×Y=\bigcup_{k=1}^m \bigcup_{j∈F_{y_k}} U_j×V_j$. Indeed consider an arbitrary element $(x, y)$ in the product. Then there exists $k∈\{1,…, m\}$ such that $y∈V_{y_k}$. In particular $y∈V_j$ for all $j∈F_{y_k}$. On the other hand $X×\{y_k\}⊆\bigcup_{j∈F_{y_k}} U_j×V_j$, whence there exists $j_0∈F_{y_k}$ such that $x∈U_{j_0}$. It follows that $(x, y)∈U_{j_0}×V_{j_0}$. ◻

Note. Fix $y∈X, i:X→X×\{y\}$ is continuous so $X×\{y\}⊆X×Y$ is compact.

$X×\{y\}$ is covered by 𝒰 so $∃J_y⊆J$ finite such that $X×\{y\}⊆\{U_j×V_j:j∈J_y\}$

May assume $y∈V_j ∀j∈J_y$ [otherwise throw it out]

Let $V_y=\bigcap_{j∈J_y} V_j$, then $y∈V_y$, $V_y$ open.

Remark. $X×V_y⊆\bigcup_{j∈J_y} U_j×V_j$

Proof. $X×\{y\}⊆\bigcup_{j∈J_y} U_j×V_j$ So given $x∈X$

$(x, y)∈U_i×V_i$ for some $i∈J_y$.

But $V_y⊆V_i⇒X×V_y⊆U_i×V_i ∀x⇒X×V_y⊆\bigcup_{j∈J_y} U_j×V_j$ ◻

Note. $\{V_y:y∈Y\}$ is a cover of $Y$

$Y$ is compact$⇒Y⊆V_{y_1}∪⋯∪V_{y_k}$ for some $y_1,…,y_k$

Claim: $\bigcup_{k=1}^∞\bigcup_{j∈J_{y_k}} \{U_j×V_j:j∈J_{y_k}\}$ is a finite subcover of 𝒰.

Proof. Let $(x, y)∈X×Y$. Then $y∈V_{y_i}$ for some $i$.

By remark, $X×V_{y_i}∈\bigcup_{s∈J_{y_i}} \{U_s×V_s:s∈J_{y_i}\}$ so $(x, y)∈\bigcup_{s∈J_{y_i}} \{U_s×V_s:s∈J_{y_i}\}$ ◻

Remark. If $X$ is a metric space and $K⊆X$ is compact then $K$ is closed and bounded.

Proof. We proved ‘bounded’ last time.

Metric spaces are Hausdorff and compact subsets of Hausdorff spaces are closed. ◻

Corollary. Suppose $X$ is compact topological space, $Y$ is a metric space and $f:X→Y$ is continuous. Then $f(X)$ is bounded and closed in $Y$.

Proof. $f(X)$ is compact$\Rightarrow$bounded and closed. ◻

Corollary. If $X$ is compact topological space and $f:X→ℝ$ is continuous, then $f$ is bounded and attains its bounds.

Proof. $f(X)$ is compact, so it is bounded, so $f$ is bounded.

$f(X)$ is closed in ℝ so $M=\sup f(X), m=\inf f(X)$ lie in $f(X)$. ◻

Remark. If $A⊆X, B⊆Y$ are compact(in topological spaces $X, Y$) then $A×B⊆X×Y$ is compact.

Proof. $A, B$ are compact with subspace topology so $A×B$ is compact(as topological space). If $i:A×B→X×Y, i(a, b) =(a, b)$ is continuous, so $i(A×B)=A×B⊆X×Y$ is compact. ◻

Theorem. Suppose $X$ is compact, $Y$ is Hausdorff, $f:X→Y$ is continuous bijection. Then $f$ is a homeomorphism.

Proof. We need to show $f^{-1}: Y→X$ is continuous. For $K⊆X$ closed, we show $(f^{-1})^{-1}(K)=f(K)$ is closed. $X$ is compact, $K⊆X$ closed, so $K$ compact, but $f$ is continuous, so $f(K)$ compact, and a compact subset of Hausdorff space $Y$ is closed, so $f(K)$ is closed. ◻

Theorem. (Heine-Borel)Any closed bounded subset of $ℝ^n$ with the standard topology is compact.

Remark. This is not true for general metric spaces e.g. take $X$ infinite with $d(x, y)=1 ∀x \neq y$, then $X$ is closed and bounded in $X$ but not compact.

Sequential compactness

Definition. A topological space is sequentially compact if any sequence in $f$ has a convergent subsequence.

A non-empty subset $A$ of $X$ is sequentially compact if sequentially compact with subspace topology, equivalently $a_n∈A ∃a_{n_k}→a$ in $A$.

Theorem. (Bolzano-Weierstrass)In a compact topological space $X$ every infinite subset has an accumulation point.

Proof. Suppose $A⊆X$ is infinite with $A'=∅$. Then $\bar A=A∪A'=A⇒A$ is closed. $∀a∈A ∃U_a$ open such that $U_a∩A=\{a\}$

$∀x∈X∖A ∃U_x$ open such that $U_x∩A=∅$

$\{U_x:x∈X\}$ is a cover of $X$ so $∃U_{x_1}∪⋯∪U_{x_n}∪U_{a_1}∪⋯∪U_{a_k}$ covers $X$ ($x_i \not\in A, a_i∈A$) then $A=\{a_1,…,a_k\}$ finite. contradiction. ◻

Theorem. A compact metric space is sequentially compact.

Proof. Seen in Metric Spaces. ◻

Theorem. Any compact metric space is complete.

Proof. Let $(x_n)$ be Cauchy. If it is finite, certainly converges. If it is infinite, by Bolzano-Weierstrass, $∃x_{n_k}→x∈X$. Claim: $\lim_{n→∞} x_n=x$.

Given $ε$. $∃k_0 ∀k⩾k_0:d(x_{n_k}, x)<\frac{ε}{2}$

$∃N ∀n, m>N:d(x_n, x_m)<\frac{ε}{2}$

Take $k>k_0$ such that $n_k>N$.

Then $∀n>N:d(x_n, x)⩽d(x_n, x_{n_k}) + d(x_{n_k}, x)<\frac{ε}{2} + \frac{ε}{2}=ε$. ◻

Theorem. Any sequentially compact metric space is compact.

Proposition. (Lebesgue lemma)Let $X$ be sequentially compact metric space. ∀ open cover 𝒰 of $X$, $∃ε>0$ such that $∀x∈X:B(x, ε)⊆U$ for some $U⊆𝒰$.

Proof. By contradiction, say no $ε$ works, so $\frac{1}{n}$ doesn't work $∀n$.

$⇒∃x_n:B \left( x_n, \frac{1}{n} \right)$ is not contained in any set of 𝒰.

$∃U∈𝒰$ contains $x$, $U$ open$⇒∃ε:B(x, 2 ε)⊆U$

$X$ is sequentially compact$⇒∃x_{n_k}→x⇒∃ε>0$ such that $∀n⩾N:x_{n_k}∈B(x, ε)$

For $k$ large enough $\frac{1}{n_k}<ε$, whence $B \left( x_{n_k}, \frac{1}{n_k} \right)⊆B(x_{n_k}, ε)⊆B(x, 2 ε)⊆U$. contradiction. ◻

$X$ metric space

Recall $X$ is sequentially compact if $∀(x_n)∃x_{n_k}:(x_{n_k})$ converges.

Theorem. $X$ is compact$⇒X$ is sequentially compact.

Proof. Seen in Metric Spaces. ◻

Theorem. $X$ is sequentially compact$⇒X$ is compact

Proof relies on 2 lemmas:

Lemma 1. (Lebesgue)Let $X$ be a sequentially compact metric space. For any open cover 𝒰 of $X$ there is some $ε>0$ such that $∀x∈X∃U∈𝒰: B (x,ε)⊂U$.

Proof. Done last time. ◻

Definition. The ε of this lemma is called a Lebesgue number of the cover 𝒰.

Definition. Given $ε>0$, an ε-net of a metric space is a set $N⊂X$ such that $\{B (n,ε):n∈N\}$ is a cover of $X$.

Example.

$ℤ^2⊂ℝ^2$ is a $\sqrt{2}$-cover of $ℝ^2$
$ℤ⊂ℝ$ is a 1-cover of $ℝ$

Lemma 2. Let $X$ be a sequentially compact metric space, and let $ε>0$. Then $\exists$ finite ε-net for $X$.

Proof. Seen in Metric Spaces. ◻

Proof of Theorem. Let 𝒰 be a cover of $X$. Since $X$ is sequentially compact by Lemma 1, 𝒰 has a Lebesgue number $ε>0$ (ie. $∀x, B (x,ε)⊂U$ for some $U⊂𝒰$)

By Lemma 2 $\exists$ finite ε-net $N=\{x_1, …,x_n\}$. So $X=B (x_1,ε)∪⋯∪B (x_n,ε), B (x_i,ε)⊂U_i∈𝒰$.

So $X=\bigcup_{i=1}^n U_i$ i.e. 𝒰 has a finite subcover$⇒X$ compact.

Quotient spaces

We use quotient spaces to

visualize spaces that are hard to see
define new spaces

Mathematician's glue

Definition. For a set $X$, a relation ℛ is a subset of $X×X$.

If $x∈X$, denote by $[x]$ its equivalence class.

$R$ is an equivalence relation if $xℛx ; xℛy⇒yℛx ; xℛy, yℛz⇒xℛz$

Remark. Giving a partition of a set $X$ is equivalent to defining an equivalence relation on $X$.

Let $(X, 𝒯)$ be a topological space and let $ℛ⊂X×X$ be an equivalence relation on $X$.

Definition. The set of equivalence classes of ℛ is denoted by $X/ℛ$ and is called the quotient space of $X$ with respect to ℛ.

The map $p:X→X/ℛ$ is called the collapsing map.

Proposition. Let $(X, 𝒯)$ be a topological space and ℛ be an equivalence relation on $X$. The family $\tilde{𝒯}$ of sets $\tilde{U}⊂X/ℛ$ such that $p^{-1}(\tilde{U})∈𝒯$ is a topology on $X/ℛ$ called the quotient topology.

Proof. (T1) $p^{-1}(∅)=∅∈𝒯, p^{-1}(X/ℛ)=X∈𝒯$. So $∅, X/ℛ∈\tilde{𝒯}$.

(T2) $p^{-1}(\tilde{U}∩\tilde{V})=p^{-1}(\tilde{U})∩p^{-1}(\tilde{V})∈𝒯⇒\tilde{U}∩\tilde{V}∈\tilde{𝒯}$

(T3) $p^{-1}\left( \bigcup_{i∈I} \widetilde{U_i} \right)=\bigcup_{i∈I} p^{-1}(\widetilde{U_i})∈𝒯⇒\bigcup_{i∈I} \widetilde{U_i}∈\tilde{𝒯}$ ◻

Remark. By definition of quotient topology, $p:X→X/ℛ$ is continuous.

Proposition. $X$ is a topological space. ℛ is an equivalence relation. $\tilde{V}⊂X/ℛ$ is closed in $X/ℛ$ iff $p^{-1}(\tilde{V})$ is closed in $X$.

Proof. $p^{-1}((X/ℛ)∖\tilde V)=X∖p^{-1}(\tilde{V})$

$\tilde{V}$ is closed$⇔(X/ℛ)∖\tilde{V}$ is open$⇔X∖p^{-1}(\tilde{V})$ is open$⇔p^{-1}(\tilde{V})$ is closed. ◻

Example.

$[0,1]∪[1, 2]/ℛ ≅[0, 2]$. Let $1ℛ2$ and $xℛx∀x$
$X=[0,1], 0∼1, X/\mmlToken{mi}∼≅S^1$

Proposition 1. Let $X, Z$ be topological spaces and let ℛ be an equivalence relation of $X$. Let $g:X/ℛ→Z$ be a function. Then $g$ is continuous iff $g∘p:X→Z$ is continuous.

Proof. ($⇒$) $g$ is continuous, $p$ is continuous$⇒g∘p$ is continuous

($⇐$) Assume $g∘p$ is continuous. Let $U⊂Z$ be open.

$p^{-1}(g^{-1}(U))=(g∘p)^{-1}(U)⇒p^{-1}(g^{-1}(U))$ is open
By definition of quotient space, $g^{-1}(U)$ open$⇒g$ is continuous ◻

Back to example $[0,1]/\mmlToken{mi}∼≅S^1$

Define $g:[0,1]/\mmlToken{mi}∼→S^1$ by $g ([t])=e^{2πit}$

$0∼1$ but $g (0)=e^{2πi 0}=e^{2πi 1}=g (1)$ so well-defined

$g∘p:[0,1]→S^1$

$g∘p (t)=e^{2πit}$ which is continuous so by Prop 1 $g$ is continuous.

$g$ is 1-1, onto

$[0,1]$ is compact, so its image $p ([0,1])=[0,1]/\mmlToken{mi}∼$ is compact.

$S^1⊂ℝ^2$ is Hausdorff$⇒S^1$ is Hausdorff

Recall: If $f:X→Y$ is bijective, continuous, $X$ is compact, $Y$ is Hausdorff, then $f$ is homeomorphism.

So $[0,1]/\mmlToken{mi}∼≅S^1$.

Proposition 2. Let $f:X→Y$ be a surjective continuous map betwen topological spaces. Let ℛ be an equivalence relation on $X$ defined by the partition $\{f^{-1}(y):y∈Y\}$

If $X$ is compact, $Y$ is Hausdorff, then $X/ℛ, Y$ are homeomorphic.

Proof. Define $g:X/ℛ→Y$ by $g ([x])=f (x)$

$g$ is well-defined by definition of ℛ
$x∼y⇒f (x)=f (y)$
$g$ is onto since $f$ is onto
$g$ is 1-1 since $g ([x])=g ([y])⇒y∈f^{-1}(x)⇒x∼y$
$X/ℛ= p (X)⇒X/ℛ$ is compact
$g ([x])=g (p (x))=f (x)$ is continuous$⇒g$ is continuous by Prop 1

$Y$ is Hausdorff, $g:X/ℛ→Y$ is continuous bijection, so $X/ℛ ≅Y$. ◻

Proposition. $f:X→Y$ continuous surjection, $X$ compact, $Y$ Hausdorff, ℛ is given by the partition $X=\bigcup_{y∈Y}f^{-1}(y)$. Then $X/ℛ≅Y$.

Definition. The 2-dimensional torus $𝕋^2$ is the topological space $S^1×S^1$.

Example 1.

$X=I×I,I=[0,1]$

Let ℛ be the equivalence relation given by equivalence classes

$\{(x,0),(x,1)\}∀x∈(0,1)$

$\{(0,y),(1,y)\}∀y∈(0,1)$

$\{(0,0),(1,0),(0,1),(1,1)\}$

$\{(x,y)\}∀x,y∈(0,1)$

Claim. $X/ℛ$ is homeomorphic to $𝕋^2$

Define $f:[0,1]×[0,1]→S^1×S^1$ by $f(x,y)=(e^{2πix},e^{2πiy})$

$f$ is continuous, onto and the partition $\{f^{-1}(y):y∈𝕋^2\}$ is the one given by ℛ.

$[0,1]×[0,1]$ is compact

$S^1×S^1$ is Hausdorff ($S^1$ is Hausdorff)

So by Proposition, $I×I/ℛ≅𝕋^2$

Example 2. Consider the unit disc $D^2=\{(x,y)∈ℝ^2:x^2+y^2≤1\}$

$ℛ$ given by equivalence classes

$\{(x,y)\}$ if $x^2+y^2<1$

$\{(x,y):x^2+y^2=1\}$

Then $D^2/ℛ≅S^2$(sphere)

We define $f:D^2→S^2$

$(x,y)=(r\cosθ,r\sinθ)$

\[f(x,y)=f(r\cosθ,r\sinθ)=(\underbrace{\cosθ\sin(πr)}_a,\underbrace{\sinθ\sin(πr)}_b,\underbrace{\cos(πr)}_c)\]

$a^2+b^2+c^2=\sin^2(πr)+\cos^2(πr)=1$

1) $f$ is onto: Given $(a,b,c)∈S^1,∃ r$ such that $c=\cos(πr)$

$a^2+b^2+c^2=1⇒a^2+b^2=1-c^2=\sin^2(πr)$

So $∃θ$ such that $(a,b)=(\cosθ\sin(πr),\sinθ\sin(πr))$

2) $f$ is continuous

3) $\{f^{-1}(v):v∈S^2\}$ is the partition defined by ℛ

4) $D^2$ is compact(closed+bounded)

5) $S^2⊆ℝ^3$, $ℝ^3$ is Hausdorff, so ${S^2}$ is Hausdorff

So by Proposition, $D^2/ℛ≅S^2$.

Describing equivalence relations with pictures(shorthand)

$𝕋^2=$

Möbius band=

Definition. Let $X$ be a set and ℛ be a relation that is reflexive and symmetric, but not necessarily transitive. Then the transitive closure of ℛ is the equivalence relation $\bar{ℛ}$ given by

\[x\barℛx'⇔\text{there is a finite sequence of points }x=x_1,x_2,…,x_n=x'\text{ such that }x_iℛx_{i+1}∀i\]

Lemma. The transitive closure of a reflexive, symmetric relation is an equivalence relation.

Proof. (Reflexive) $xℛx⇒x\barℛx$

(Symmetric) Say $x=x_1,x_1ℛx_2,x_2ℛx_3,…,x_{n-1}ℛx_n,x_n=y$, ℛ is symmetric

So $yℛx_{n-1},x_{n-1}ℛx_{n-2},…,x_2ℛx_1⇒yℛx$

(Transitive) Say $x\barℛy,y\barℛz$

$x=x_1,x_1ℛx_2,…,x_{n-1}ℛx_n,x_n=y=y_1,y_1ℛy_2,…,y_{n-1}ℛy_n,y_n=z⇒x\barℛz$ ◻

Example. Klein bottle

Pathologies

Example. $X=ℝ$ with standard topology

Define $x∼y⇔x-y∈ℚ$

Denote the quotient space $ℝ/∼$ by $ℝ/ℚ$

What is the quotient topology?

Say $U⊆ℝ/ℚ$ is open. Then $p^{-1}(U)⊆ℝ$ is open. So $p^{-1}(U)⊇(a,b),a≠b$.

Note if $q∈ℚ,x∈(a,b),x+q∼x$, so $p(x+q)=p(x)$.

$⇒x+q∈p^{-1}(U)∀q∈ℚ$

$⇒p^{-1}(U)⊇\bigcup_{q∈ℚ}((a,b)+q)=ℝ$

So the topology is the indiscrete topology.

Definition. Let $X$ be a topological space and ℛ an equivalence relation on $X$. Then $A⊆X$ is saturated (wrt ℛ) if it is a union of equivalence classes.

Remark. Let ℛ be an equivalence relation on $X$ and let $p:X→X/ℛ$ be the collapsing map. TFAE

$A$ is saturated.
If $A∩[x]≠∅$ for some $x$ then $[x]⊆A$.
If $x∈A,y∼x$, then $y∈A$.
$A=p^{-1}(p(A))$

Lemma. There are 1-1 correspondences:

subsets of $X/ℛ$	⟷	saturated subsets of $X$
open subsets of $X/ℛ$	⟷	open saturated subsets of $X$
closed subsets of $X/ℛ$	⟷	closed saturated subsets of $X$

Proof. $B⊆X/ℛ⟷p^{-1}(B)⊆X$

This is 1-1 and $p^{-1}(B)$ is saturated

Conversely, if $C⊆X$ is saturated then $p^{-1}(p(C))=C$, so the map is onto.

By definition of quotient topology:

$U$ is open$⇔p^{-1}(U)$ open

$K$ is closed $⇔p^{-1}(K)$ is closed. ◻

Proposition. Let $X$ be a topological space and $X/ℛ$ be a quotient space. Then $X/ℛ$ satisfies the first separation axiom if every equivalence class in $X$ is closed.

Proof. The first separation axiom is equivalent to the statement that every singleton $\{x\}$ in $X/ℛ$ is closed$⇔p^{-1}(\{x\})$ is closed in $X$.

but $p^{-1}(\{x\})$ is an equivalence class. ◻

Corollary. A quotient space $X/ℛ$ is Hausdorff iff any 2 distinct equivalence classes are contained in disjoint open saturated sets in $X$.

Example. Klein bottle is Hausdorff.

$∀x,y∃$two open disjoint saturated sets $U∋x,V∋y$ [draw a picture]

Corollary. A quotient space $X/ℛ$ is Hausdorff iff any 2 distinct equivalence classes are contained in disjoint open saturated sets in $X$.

Example 1. $ℝ$, $xℛy⇔x-y∈ℤ$ Then $ℝ/ℤ$ is Hausdorff

Proof. Let $[a]≠[b]∈ℝ/ℤ$.

Let $ε=\min\{|a-b+n|:n∈ℤ\}$

Let $δ=\fracε2$, $U=\bigcup_{n∈ℤ}(a-δ, a+δ)+n, V=\bigcup_{n∈ℤ}(b-δ, b+δ)+n.$

$U∋[a], V∋[b]$ are disjoint, open, saturated$⇒ℝ/ℤ$ is Hausdorff ◻

Definition. The real $n$-dimensional projective space $ℝP^n$ is the quotient space of $ℝ^{n+1}∖\{0\}$ with respect to the relation $xℛy⇔∃λ≠0, λ∈ℝ$ such that $y=λx$

ie. $ℝP^n$ is the set of lines through 0.

($ℝ^{n+1}∖\{0\}$ standard topology, $ℝP^n$ quotient topology, open sets in $ℝP^n$ are open cones)

Example 2. $ℝP^n$ is Hausdorff.

Let $[a], [b]∈ℝP^n$. Consider the points $\frac{a}{‖a‖}=a_1,-\frac{a}{‖a‖}=a_2, \frac{b}{‖b‖}=b_1,-\frac{b}{‖b‖}=b_2,$

If $S^n=\{x∈ℝ^{n+1}:‖x‖=1\}$ then $[a]∩S^n=\{a_1, a_2\}, [b]∩S^n=\{b_1, b_2\}$

Let $ε=\min\{‖a_i-b_i‖:i=1, 2, j=1, 2\}$. Let $δ=\fracε2$. Consider $(B(a_1,δ)∪B(a_2,δ))∩S^n=U,(B(b_1,δ)∪B(b_2,δ))∩S^n=V$

$U, V$ are open in $S^n$

Let $f:ℝ^n∖\{0\}→S^n$ be $f(x)=\frac{x}{‖x‖}$

$f$ is continuous, so $\tilde{U}=f^{-1}(U), \tilde{V}=f^{-1}(V)$ are disjoint, saturated subsets of $ℝ^{n+1}∖\{0\}$ such that $[a]∈\tilde{U}, [b]∈\tilde{V}$.

So $ℝP^n$ is Hausdorff.

[Why saturated? If $x∈\tilde{U}, f(x)=\frac{x}{‖x‖}∈U⇒\frac{λx}{\|λx‖}∈U$, so $λx∈\tilde{U}∀λ∈ℝ∖\{0\}$, so $[x]⊂\tilde{U}⇒\tilde{U}$ saturated]

Notation. $ℝ^3 \supset S^2=\{(x, y, z)∈ℝ^3:x^2+y^2+z^2=1\}$

$D^+=\{(x, y, z)∈S^2:z≥0\}$

$D^2=\{(x, y)∈ℝ^2:x^2+y^2≤1\}$

Proposition. The following are all homeomorphic to the projective plane $ℝP^2$

a) $S^2/ℛ$ where $(x, y, z)ℛ(-x,-y,-z)∀(x,y,z)∈S^2$

b) $D^+/ℛ$ where $(x, y, 0)ℛ(-x,-y,0)∀(x, y, 0)∈D^+$

c) $D^2/ℛ$ where $(x,y)ℛ(-x,-y)$ if $x^2+y^2=1$

Proof. a) Restrict the collapsing map $p:ℝ^3∖\{0\}→ℝP^2$ to $S^2$ to get $f:S^2→ℝP^2$

$f$ is continuous and surjective.

The equivalence relation given in (a) is given by the partition $\{f^{-1}(y):y∈ℝP^2\}$

$S^2$ is compact, $ℝP^2$ is Hausdorff.

By Prop last time, $S^2/ℛ≅ℝP^2$ is homeomorphism.

b) same proof except restrict $p$ to $D^+$

c) The map $g:D^+→D^2$

$g (x, y, z)=(x, y)$

or $g(x,y,z)=(\frac x{z+1},\frac y{z+1})$ [steorographic projection from the south pole] is 1-1 onto continuous. $D^+$ compact, $D^2$ Hausdorff$⇒g$ is homeomorphism.

So $f∘g^{-1}:D→ℝP^2$ is continuous and reduces ℛ to $D^2$ as before.

So $D^2/ℛ≅ℝP^2$.

d) $D^2≅[0,1]×[0,1]$ and use c) ◻

Definition. A function $f:X→Y$ between topological spaces is an open mapping if $∀U⊂X$ open $f(U)$ is open in $Y$

Remark. 1) $f:X→Y$ continuous bijective and open$⇔f$ is a homeomorphism

2) $p:[0, 1]→[0, 1]/\small0∼1$ is not open

Eg. $p \left( \left[ 0, \frac{1}{2} \right) \right)$ is not open

Quotient maps

Proposition. Let ℛ be the equivalence relation on topological space $X$ and let $p:X→X/ℛ$ be given by $p (x)=[x]$. Then the quotient topology on $X/ℛ$ is the finest topology that makes $p$ continuous.

Proof. Let $𝒯'$ be a topology on $X/ℛ$ such that $p$ is continuous then if $U∈𝒯', p^{-1}(U)$ is open.

So $U∈𝒯$(quotient topology), so $𝒯'⊂𝒯$. ◻

Definition. A map $f:X→Y$ between topological spaces is a quotient map if

1) $f$ is surjective

2) $∀U⊂Y, U$ is open$⇔f^{-1}(U)$ is open

Remark. If $f:X→Y$ is surjective, continuous and open, then $f$ is a quotient map.

Proposition. Let ℛ be an equivalence relation on a space $X$ and equip $X/ℛ$ with quotient topology. Then $p:X→X/ℛ, p (x)=[x]$ is a quotient map.

Proof. By definition of quotient map, $U$ is open$⇔p^{-1}(U)$ is open in $X/ℛ$. ◻

Proposition. Suppose $q:X→Y$ is a quotient map and ℛ is the equivalence relation given by the partition $\{q^{-1}(y):y∈Y\}$. Then $X/ℛ$ is homeomorphic to $Y$.

Proof. Consider $g:X/ℛ→Y$ given by $g ([x])=q (x)$. Then $g$ is onto since $q$ onto. 1-1 by definition of ℛ.

Recall $g$ continuous$⇔g∘p=q$ continuous (past Proposition)

So $g$ is continuous.

To show $g$ is a homeomorphism, it is enough to show that $g$ is open.

Let $U⊂X/ℛ$ be open.

$g (U)⊂Y, q$ is quotient map, so $g (U)$ open$⇔q^{-1}(g (U))$ open

But $q^{-1}(g (U))=(g∘p)^{-1}(g (U))=p^{-1}(\underbrace{g^{-1}(g (U))}_U)=p^{-1}(U)$ open by definition of quotient topology. So $g (U)$ is open. ◻

If $q:X→Y$ is a quotient map and $ℛ$ is given by the partition $\{q^{-1}(y):y∈Y\}$ then $X /ℛ≃y$

Definition. $f:X→Y$ is a quotient map if $f$ is surjective and $U⊂Y$ is open iff $f^{-1}(U)$ is open.

Remark. If $f:X→Y$ is continuous surjective open, then $f$ is a quotient map.

Example. $ℝ/ℤ$ is homeomorphic to $S^1$

Proof. $f:ℝ→S^1, f (x) =e^{2πix}$

$x∼y⇔f (x) =f (y)⇔e^{2πi (x-y)} =1⇔x-y∈ℤ$

So the equivalence relation of $ℝ/ℤ$ is given by $\{f^{-1}(x):x∈S^1\}$

$f$ is surjective, continuous.

Also $f$ is open, since $f ((x-ε, x+ε))$ is an open arc on $S^1$.

So by Proposition, $ℝ/ℤ≃S^1$. ◻

Example. Define ∼ on $ℝ^n$ by $x∼y⇔x-y∈ℤ^n$. Then $ℝ^n /ℤ^n≃𝕋^n =\underbrace{S^1×⋯×S^1}_n$

Proof. $F (x_1,…x_n) =(e^{2πix_1},…e^{2πix_n})$

$F$ continuous, surjective, reduces ∼ and open, so $F$ is a quotient map. ◻

Geometric Topology

Objective. Study “nice” spaces that appear in geometry up to homeomorphism.

Create a “list” of these nice spaces
The spaces will be built by gluing simplices
We'll give a “finite”, “combinatorial” description of these spaces
Convince ourselves that nice spaces appear (up to homeomorphism) in our list
Show some basic properties
Classify up to homeomorphism all surfaces

Simplicial complexes

Definition. The standard $n$-simplex is the set

\[Δ^n =\left\{(x_1,…x_{n+1})∈ℝ^{n+1}:x_i≥0∀i \text{ and } \sum_{i =1}^{n+1} x_i =1 \right\}\]

The non-negative integer $n$ is the dimension of the simplex.

Its vertices denoted by $V (Δ^n)$ are points $(x_1,…x_{n+1})$ where some $x_i =1$, all the others are 0.

$∀A \subseteq \{1,…n+1\}$ there is a corresponding face of $Δ^n$ which is

\[\{(x_1,…x_{n+1})∈Δ^n:x_i =0 \text{ if } i \not\in A\}\]

Eg. $Δ^1$ has 3 faces.

The inside of a simplex is

\[\operatorname{Inside} (Δ^n) =\left\{(x_1,…x_{n+1})∈ℝ^{n+1}:x_i > 0∀i \text{ and } \sum_{i =1}^{n+1} x_i =1 \right\}\]

Remark. The vertices $\{v_1,…v_{n+1}\}$ of the standard simplex $Δ^n⊂ℝ^{n+1}$ are the basis elements of the standard basis of $ℝ^{n+1}$. If $\{e_1,…e_{k+1}\}$ is standard basis of $ℝ^{k+1} (k<n)$ and we define $T (e_i) =V_{n_i}$ where $V_{n_1},…V_{n_{k+1}}$ are all distinct, then $T$ extends uniquely to a 1-1 linear map $T:ℝ^{k+1}→ℝ^{n+1}$ and $T (Δ^k)$ is a face of $Δ^n$.

Definition. A face inclusion of a standard $m$-simplex $Δ^m$ into a standard $n$-simplex $Δ^n (m<n)$ is a function $Δ^m→Δ^n$ that is a restriction of an injective linear map that sends the vertices of $Δ^m$ to vertices of $Δ^n$.

Note. Any 1-1 map $V (Δ^m)→V (Δ^n)$ extends to a unique face inclusion.

Eg. For $Δ^1→Δ^2$, $V (Δ^1) =\{1, 2\}→V (Δ^2) =\{1, 2, 3\}$ have 6 possible maps:

Definition. An abstract simplicial complex is a pair $(V, Σ)$ where $V$ is a set called vertices and $Σ$ is a set of non-empty subset of $V$ called set of simplices such that

$∀v∈V, \{v\}∈Σ$
If $σ∈Σ$, then any non-empty $τ⊂σ$ lies in $Σ$ as well.

We say $(V, Σ)$ is finite if $V$ is finite.

Definition. The topological realization $|K|$ of an abstract simplicial complex $K =(V, Σ)$ is the space obtained by the following procedure:

$∀σ∈Σ$, if $|σ|=n+1$, take a copy of $Δ^n$.
Denote this by $Δ_σ$ and label its vertices with elements of $σ$.
whenever $σ⊂τ∈Σ$ identify $Δ_σ$ with a subset of $Δ_{τ}$ via the face inclusion that sends the elements of $σ$ to the corresponding elements of $τ$.

In other words, $|K|=\bigsqcup_{σ∈Σ} Δ_σ /ℛ$ where $ℛ$ is described in 2.

Whenever we refer to a simplicial complex, we will mean either an abstract simplicial complex or its topological realisation.

Example. $V =\{1, 2, 3\}, Σ =\{\{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}\}$　

Example. $V =\{1, 2, 3, 4\}, Σ =\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2, 3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{3, 4\}\}$

Example. is Not a simplicial complex. $\{1, 2\}$ gives one simplex

Example. is Not a simplicial complex. $\{1, 2\}$ gives a unique simplex.

Definition. A triangularization of a space $X$ is a simplicial complex $K$ together with a homeomorphism $f:|K|→X$. (Note: not unique!)

Example. is $|K|$

$∃ϕ:|K|→S^2$ homeomorphism

Example. Triangularization of the torus $𝕋^2$

At first sight, this triangulation of the torus may seem be needlessly complicated. Might we have been able to use fewer simplices?

andare not allowed because $Δ^2 =\{1, 2\}$ must determine a single edge

Definition. A simplicial circle is a simplicial complex $K$ with vertices $\{v_1,…v_n\}$ and 1-simplices $\{v_1, v_2\}, \{v_2, v_3\},…\{v_n, v_1\}$

Eg.

Lemma. The topological realization of a simplicial complex $K=(V,Σ)$ is the union of the insides of its simplices. Moreover, for distinct simplices, their insides are disjoint, ie. $|K|=\bigsqcup_{σ∈Σ}\text{inside}(Δ_σ)$

Proof. $|K|=\bigsqcup_{σ∈Σ}Δ_σ/ℛ$ where $τ⊂σ$, ${Δ_{τ}}$ is identified with a face of $Δ_σ$.

$X∈\text{inside}(Δ_σ)⇔X$ does not lie in a proper face of $Δ_σ$ or $Δ_σ$ is a vertex.

Let $x∈|K|$. Then $x∈Δ_σ$ for some $σ$. Take $σ$ with smallest cardinality such that $x∈Δ_σ$. Then $X∈\text{inside}(Δ_σ)$.

By definition of $ℛ$, points $x∈\text{inside}(Δ_σ),y∈\text{inside}(Δ_{τ}){,}τ⊆σ$ are never equivalent. ◻

Definition. A subcomplex of a simplicial complex $K=(V,Σ)$ is a simplicial complex $K'=(V',Σ')$ where $V'⊂V,Σ'⊂Σ$.

Lemma. If $K'$ is a subcomplex of $K$, then $|K'|$ is a closed subset of $|K|$.

Proof. Let $p:\bigsqcup_{σ∈Σ}Δ_σ→|K|$ be the collapsing map. Then $p^{-1}(|K'|)$ is a union of faces of the $Δ_σ$'s. But a face of $Δ_σ$ is closed in $Δ_σ$, so $p^{-1}(|K'|)$ is closed. By definition of quotient topology, $X⊂|K|$ is closed iff $p^{-1}(X)$ is closed, so $|K'|$ is closed. ◻

Definition. For a simplicial complex $K=(V,Σ)$ and $V'⊆V$ the subcomplex spanned by $V'$ has vertex set $V'$ and consists of all simplices that have all their vertices in $V'$.

Definition. Let $V=(K,Σ)$ be a simplicial complex and let $v∈V$ be a vertex.

The link of $v$, denoted by $\operatorname{lk}(v)$, is a subcomplex with vertex set $\{w∈V ∖\{v\}:\{v,w\}∈Σ\}$
and simplices $σ$ such that $v∉σ$ and $σ∪\{v\}∈Σ$
The star of $v$, denoted $\operatorname{st}(v)$ is $\bigcup\{\operatorname{inside}(σ):σ∈Σ\text{ and }v∈σ\}$

Lemma. The star of a vertex $v$ is an open set containing $v$.

Proof. We will show ${|K|}∖\operatorname{st}(v)$ is closed.

Claim. ${|K|}∖\operatorname{st}(v)=|\operatorname{span}(V∖\{v\})|$

Proof. If $x∈{|K|}∖\operatorname{st}(v)$ then $∃σ$ such that $x∈\operatorname{inside}(Δ_σ)$ and $v∉σ,$ so $x∈\operatorname{span}(V ∖\{v\})$. Conversely, if $x∈\operatorname{span}(V ∖\{v\})$, then $x∈Δ_σ$, $Δ_σ⊂{|\operatorname{span}(V ∖\{v\})|}$, so $x$ lies in the inside of some face of $Δ_σ$, so $x∉\operatorname{st}(v)$. ◻

Since any subcomplex is closed, $|\operatorname{span}(V∖\{v\})|$ is closed, so $\operatorname{st}(v)$ is open. ◻

Elementary properties of simplicial complexes.

Definition. An edge path of a simplicial complex is a sequence of vertices $X_1,…,X_n$ such that $\{X_i,X_{i+1}\}∈Σ∀i=1,…,n-1$.

We say that $K$ is edge path connected if $∀v,w∈V∃$edge path from $v$ to $w$.

Proposition. Let $K=(V,Σ)$ be a simplicial complex. Then the following are equivalent:

$K$ is edge path connected.
$|K|$ is path connected.
$|K|$ is connected.

Proof. 1)⇒2) Suppose that any two vertices in $K$ can be joined by an edge path. Let $x$ and $y$ be any two points in $|K|$. Each lies in a simplex, and so there is a straight line in that simplex joining it to a vertex. These two vertices can be joined by an edge path, which can be realised as a path in $|K|$. So, $|K|$ is path-connected.

2)⇒3) path connected⇒connected

3)⇒1) Suppose $|K|$ is not edge path connected. Let $V_1=\{w:v,w\text{ can be joined by an edge path}\}$. Let $V_2=V ∖ V_1$. Then $|K|=|\operatorname{span}(V_1)|⊔|\operatorname{span}(V_2)|⇒|K|$ union of disjoint closed sets$⇒|K|$ is not connected. ◻

Lemma. The topological realisation of a finite simplicial complex $K=(V,Σ)$ is compact.

Proof. $p:\bigsqcup_{σ∈Σ}Δ_σ→|K|$ continuous. Since $\bigsqcup_{σ∈Σ}Δ_σ$ is compact, $|K|$ is compact. ◻

Proposition. For any finite simplicial complex $K=(V,Σ)$ there is a continuous injection $f:|K|→ℝ^n$ for some $n$.

Proof. Let $K'=(V,Σ'),Σ'=\{σ:σ∈V\}$

Then $|K'|$ is a simplex, with $|V|=k$ vertices. So $|K'|⊆ℝ^{k+1}$. Also $K$ is a subcomplex of $K'$, so $i:|K|\hookrightarrow|K'|$ is continuous 1-1. ◻

Corollary. The topological realisation of a simplicial complex is metrizable.

Proof. $f:|K|→ℝ^n$ is 1-1, continuous, $|K|$ compact, $ℝ^n$ Hausdorff. ${|K|}≅f({|K|})⊂ℝ^n$ is a metric space. ◻

Corollary. $|K|$ is Hausdorff.

Proof. metrizable⇒Hausdorff. ◻

Definition. An $n$-dimensional manifold is a Hausdorff topological space $M$ such that any point of $M$ lies in an open set homeomorphic to an open set in $ℝ^n$. We often simply say n-manifold.

Example. $ℝ^n$,

$S^n$(since $S^n ∖\{p\}≈ℝ^n$)

$\mathbb{T}^n,ℝP^n$

Remark. An open subset of an n-manifold is an n-manifold. These are often “wild”, we will focus on compact manifold (called closed). eg. $S^n,ℝP^n$ etc.

Definition. A 2-dimensional manifold is called a surface.

Goal: Classify closed surfaces (=compact 2-dimensional manifolds)

Eg. $S^2,\mathbb{T}^2,ℝP^2$

Polygons with a complete set of side identification (CSSI)

Let $P$ be a finite sided convex polygon with an even number of sides. We arrange the side in pairs and identify (using quotient topology) the sides in a pair.

Precisely if $e,e^1$ is a pair of edges with vertices $v_1,v_2$ and $v_1',v_2'$ respectively.

[or $tv_1+(1-t)v_2∼tv_1'+(1-t)v_2'$]

Then take the quotient space $P/∼$

Example. torus $aba^{-1}b^{-1}$

Klein bottle $aba^{-1}b$

$ℝP^2$ $abab$

Proposition. A polygon with a CSSI is a closed surface.

Proof. $P$ is clearly homeomorphic to a regular polygon. So we assume $P$ is regular. $P$ is compact, so $P/\mmlToken{mi}∼$ is compact. Need to show that $S=P/\mmlToken{mi}∼$ is a surface. There are 3 types of points

A point $x$ in the interior of $P$. Take a small ball $B(x,ε)$ in the inside of $P$. Then if $π:P→P/\mmlToken{mi}∼=S$ is the collapsing map, $π(B(x,ε))$ is homeomorphic to $B(x,ε)$, since $B(x,ε)↦π(B(x,ε))$ is 1-1, onto, continuous, open.
A point $x$ on a side. Take $B(x_1,ε)∪B(x_2,ε)/\mmlToken{mi}∼$ is an open disc around $[x_1]$ in $S$. [Note that one reason that we arranged for $P$ to be regular was so that these two discs patch together correctly.]
A vertex $x$ on $S$, so $[v]=\{v_1,v_2,…,v_n\}$, $B(v_i,ε)$ is a sector
Note. angle doesn't matter for homeomorphism.
$⋃B(v_i,ε)/\mmlToken{mi}∼$
Pizza slices which we identify successively to get an open disc around $[v]$ in the quotient.

This verifies the main condition in the definition of a surface. But we still need to check that $S$ is Hausdorff. It is clear that, given two distinct points y and y′ in $S$, their open neighbourhoods described above are disjoint, provided ε is small enough. ◻

Two list of surfaces

We describe a polygon of CSSI by a word. We fix a vertex and run around the boundary of $P$ say in the clockwise direction. If $e,e'$ get identified, associate them with the same letter. We use inverses to indicate the direction of the edge.

First list. $M_g(g⩾0)$ is called $g$-holed torus

$M_0=XX^{-1}YY^{-1}$

$M_g(g⩾1)=X_1 Y_1 X_1^{-1}Y_1^{-1}…X_g Y_g X_g^{-1}Y_g^{-1}$

Remark. $M_0≅S^2$

Second list. $N_h(h⩾1)$ is called surface with $h$ crosscaps

$N_1=XXYY^{-1}$

$N_h(h⩾2)=X_1 X_1…X_h X_h$
Remark. $N_1≅ℝP^2$. Klein bottle $aba^{-1}b$ homeomorphic to the surface with 2 crosscaps [by cutting along the diagonal of the square]

Adding handles and crosscaps

Definition. Let $S$ be a surface and let $T$ be a torus. Let $D_1⊂S$, $D_2⊂T$ be homeomorphic to closed discs. Let $C_1,C_2$ be respectively their boundary circles. We pick $φ:C_1→C_2$ a homeomorphism. Let $S'=S∖\mathring{D_1},T'=T∖\mathring{D_2}$. Define $M=S'⊔T'/{\small x∼φ(x)∀x∈C_1}$

We say that $M$ is obtained from $S$ by adding a handle.

Remark. $M$ is a surface and does not depend on choice of $D_1,D_2$, $φ$(but we don't prove this)

Proposition. Let $S$ be a surface obtained from a polygon with a CSSI. Let $A$ be the boundary word. If $x,y,x^{-1},y^{-1}$ do not appear in $A$, then the word $Axyx^{-1}y^{-1}$ gives a surface that is obtained from $A$ by adding a handle.

Proof. Let $P$ be the polygon used to build $S$, and let $P'$ be the polygon with boundary word $Axyx^{-1}y^{-1}$. Let $u$ and $v$ be the points on $P'$ at the start and end of the word $A$. Note that the identifications specified by gluing the edges $x$ to $x^{-1}$ and $y$ to $y^{-1}$ result in gluing $u$ to $v$. Let $R$ be the straight arc in $P'$ running from $u$ to $v$. This arc divides $P'$ into two polygons $P_1$ and $P_2$.

Consider the polygon $P_1$. Let us glue its edges together according to the recipe given by $A$. Let us also identify the two vertices which lie at the endpoints of $R$. The resulting space $S'$ is homeomorphic to $S$ with the interior of a closed disc removed. In this quotient space, the endpoints of the arc $R$ have been glued up, to form a circle $C_1$.
The polygon $P_2$ has five sides. When the four labelled sides are glued together, the result is a torus with the interior of a closed disc removed. Call this space $T'$. The remains of the arc $R$ form a circle $C_2$. When the copy of $R$ in $P_1$ is glued to the copy of $R$ in $P_2$, this induces a homeomorphism $ϕ:C_1→C_2$. If we glue $S'$ to $T'$ using $\phi$, the resulting space is obtained from $S$ by adding a handle. This is the space that is obtained from $P'$ by making the side identifications using the boundary word $A x y x^{-1} y^{-1}$. ◻

Corollary. The surface $M_g$ is obtained from the sphere by adding $g$ handles.

Lemma. $ℝP^2$ is obtained from a Möbius band and a disc by identifying their boundary circles via a homeomorphism.

Proof. $ℝP^2$ is obtained from the unit disc $D$ by identifying each point $(x,y)$ on the boundary of the disc with $(-x,-y)$. Let ∼ be the equivalence relation. Let $A$ be the annulus $\{(x,y)∈ℝ^2:1/4⩽x^2+y^2⩽1\}$ and let $C$ be the inner boundary curve $\{(x,y)∈ℝ^2:x^2+y^2=1/4\}$. If one attaches a disc to $A$ along the curve $C$, the result is $D$. So, if one attaches a disc to the copy of $C$ in $A/∼$, the result is $ℝP^2$.

Therefore, to prove the lemma, it suffices to show that $A/∼$ is homeomorphic to the Möbius band. A proof is summarised in the following figure. ◻

Last time defined $M_g,N_h$

Definition. Let $S$ be a closed surface, and let $D_1⊂S,D_2⊂ℝP^2$ be closed discs. Denote $C_1=∂D_1,C_2=∂D_2$. We say that $M$ is obtained from $S$ by adding a crosscap if $M$ is obtained from $S∖\mathring{D_1}$ and $ℝP^2∖\mathring{D_2}$ by identifying $C_1,C_2$ via a homeomorphism.

Proposition. Let $S$ be obtained from a polygon with a CSSI. Let $A$ be its boundary word. Suppose that $A$ does not contain the letters $x,x^{-1}$. Then the word $Axx$ gives us a surface obtained from $S$ by adding a crosscap.

Corollary. $N_h$ is obtained from the 2-sphere by adding $h$ crosscaps

Proof. $N_1=ℝP^2$ is obtained from $S^2$ by adding a crosscap.

$N_2=xxyy=N_1 yy$ is obtained from $N_1$ by adding a crosscap … ◻

Fact. All compact 2-manifolds can be triangulated. [The proof is beyond our scope]

Definition. A closed combinatorial surface is a connected, finite simplicial complex $K$ such that for every vertex $v$ of $K$, $\operatorname{lk}(v)$ is a simplicial circle.

Lemma 5.16. Let $K$ be a closed combinatorial surface. Then

Every simplex of $K$ has dimension 0,1,2
Every 1-simplex of $K$ is a face of exactly two 2-simplices
Every point of $|K|$ lies in a 2-simplex
$|K|$ is a closed surface

Proof.

Let $\{v_1,…,v_n\}$ be a simplex. $\{v_2,…,v_n\}⊂\operatorname{lk}(v_1),\operatorname{lk}(v_1)$ is a simplicial circle$⇒\{v_2,…,v_n\}$ is 0 or 1 simplex, so $n=0,1,2$.
Let $\{v_1,v_2\}$ be a 1-simplex. Then $v_2∈\operatorname{lk}(v_1),\operatorname{lk}(v_1)$ is a simplicial circle, in which $v_2$ has exactly 2 adjacent vertices say $v_3,v_4$. So the simplices $\{v_1,v_2,v_3\},\{v_1,v_2,v_4\}$ are exactly the simplices containing $\{v_1,v_2\}$.
Every $x∈{|K|}$ lies in the inside of a simplex $Δ_σ$ of dimension 0,1,2 (by 1)
If $Δ_σ$ is a 2-simplex, done;
If $Δ_σ$ is a 1-simplex, done by 2)
If $Δ_σ=\{x\}$ is a 0-simplex, take $v∈\operatorname{lk}(x)$. Then $\{v,x\}$ lies in a 2-simplex by 2) so $x$ lies in a 2-simplex.
$|K|$ is Hausdorff [We showed earlier finite simplicial complex is Hausdorff] Let $x∈K$, we need to show $x$ has a neighbourhood homeomorphic to an open disc in $ℝ^2$
If $x∈\operatorname{inside}(Δ_σ)$ of dimension 2, then ok
If $x∈\operatorname{inside}(\text{edge})$ then by (2).
If $x$ is a vertex then $\operatorname{st}(x)$ is a simplicial circle, so $\operatorname{st}(x)$ is an open disc.

Proposition. Any polygon with a complete CSSI is homeomorphic to a closed combinatorial surfaces.

Proof. See notes

The classification theorem

Every closed combinatorial surface is homeomorphic to one of the manifolds $M_g(g⩾0)$ or $N_h(h⩾1)$.

The proof proceeds in two steps. First we show that any closed combinatorial surface is obtained as a polygon with a complete set of side identifications. Then we show how the polygon and its side identifications can be modified without changing the space up to homeomorphism. By a careful application of these modifications, we show how to change the boundary word until it is in the required standard form.

Proposition. If $K$ is a closed combinatorial surface, then $|K|$ is homeomorphic to a space obtained from a $2 n$-gon for some integer $n⩾2$ by identifying its edges in pairs.

Proof. ${|K|}=\bigsqcup_{σ∈Σ} Δ_σ/\mmlToken{mi}∼$

Since $K$ is connected, each 0-simplex $v$ lies in a 1-simplex $τ$ which by previous lemma lies in a 2-simplex $σ$. So identify $v$ to $τ$ and $τ$ to $σ$. So we see $|K|$ is obtained by identifying 2-simplices.

Pick $σ_1$ 2-simplex. If $e$ edge of $σ_1$, then $e$ lies in exactly one other 2-simplex $σ_2$. Identify ${σ_1}$ to $σ_2$ along $e$.

If some side of this polygon is identified to a side of another 2-simplex $σ_3$, glue $σ_3$ to this polygon along this edge.

Continue the same way we obtain a polygon $L$ such that every side of $L$ is identified to some other side of $L$.
Need to show that we used all 2-simplices of $|K|$.

Remark. Every edge of $L$ lies in exactly two 2 simplices of $L$ (by definition of $L$). It follows that if $v$ is vertex of $L$ then $\operatorname{lk}(v)$ is a simplicial circle.

On the other hand, $\operatorname{lk}(v)$ in $L$ is contained in $\operatorname{lk}(v)$ in $|K|$. So if $v∈L$ then $\operatorname{lk}(v)$ in $k=\operatorname{lk}(v)$ in $L$.

Back to proof that ${|K|}=L/\mmlToken{mi}∼$

Case I. All vertices of $K$ lie in $L$.

If $x∈{|K|}$ then $x$ lies in a closed simplex σ.

If $v$ is vertex of σ then $v∈L⇒\operatorname{lk}(v)∈L⇒σ⊂L$.

Case II. Some vertex $v$ of $K$ is not in $L$.

Let $p$ be a path in $K$ from $v$ to $v_1∈L$.

Let $v_2$ be the last vertex not in $L$.

If $v_3$ is adjacent to $v_2$ then $v_3∈L,v_2∈\operatorname{lk}(v_3)⇒v_2∈L$

Classification of closed combinatorial surfaces

Every surface is homeomorphic to a polygon with CSSI. Each such polygon with CSSI is encoded by a word $M_g,N_h$($g⩾0,h⩾1$) $a_1 b_1 a_1^{-1}b_1^{-1}…a_g b_g a_g^{-1}b_g^{-1}=M_g$, $a_1 a_1…a_h a_h=N_h$.

If $A=x_1…x_n$ is a word, denote by $A^{-1}$ the word $A^{-1}=x_n^{-1}…x_2^{-1}x_1^{-1}$

A cyclic permutation of $A$ is $x_n x_1…x_{n-1}$

Lemma 0. Let $P$ be a polygon with CSSI and boundary word $A$. Then a polygon with boundary word $A^{-1}$ or a polygon with boundary word a cyclic permutation of $A$ represents the same surface as $A$.

Proof. $A^{-1}$ has the same identifications read in opposite direction.

Cyclic permutation: Just changes the point where we start reading the word. ◻

Lemma 1. Let $P$ be a polygon with CSSI and boundary word $x A x B$ where $x$ is a letter and $A,B$ are non-empty words. Then the words $A^{-1}xxB$ and $xxA^{-1}B$ give the same surface.

Proof. Cut along $y$, glue along $x$. ◻

Lemma 2. Let $P$ be a polygon with CSSI and boundary word $x A B x^{-1}C$, where $x$ is a letter and $A,B,C$ are words. Then $x B A x^{-1}C$ represents the same surface.

Proof. Cut along $y$, glue along $x$. ◻

Lemma 3.The word $A x x y z y^{-1}z^{-1}B$ ($x,y,z$ are letters) and $A x x y y z z B$ represents the same surface.

Proof.

\[A x x y z y^{-1}z^{-1}B\xrightarrow{\text{Lemma 1}}A x z^{-1}y^{-1}x y^{-1}z^{-1}B\xrightarrow{\text{Lemma 1}}Axz^{-1}y^{-1}y^{-1}x^{-1}z^{-1}B\xrightarrow{\text{Lemma 2}}A\,x\,y^{-1}\,y^{-1}z^{-1}x^{-1}z^{-1}B\xrightarrow{\text{Lemma 1}}\]

◻

Corollary. 1) Klein bottle$=a\,b\,a^{-1}b\xrightarrow{\text{Lemma 1}}a\,a\,b\,b=N_2$

2) $yzy^{-1}z^{-1}xx\xrightarrow{\text{cycle}}z^{-1}xxyzy^{-1}\xrightarrow{\text{Lemma 2}}z^{-1}xyxzy^{-1}\xrightarrow{\text{cycle}}xyxzy^{-1}z^{-1}\xrightarrow{\text{Lemma 1}}xxy^{-1}zy^{-1}z^{-1}\xrightarrow{\text{cycle}}y^{-1}zy^{-1}z^{-1}xx\xrightarrow{\text{relabel}}yzyz^{-1}xx$

torus+crosscap=Klein bottle+crosscap

Lemma 4. If $A,B$ are non-empty words of total length ⩾4 then the words $A\,x\,x^{-1}B$ and $A\,B$ represent the same surface.

Proof. glue along $x,x^{-1}$ ◻

Lemma 5. Let $P$ be a polygon with a CSSI and at least 6 vertices. Let $S=P/∼$. If $P$ has more than 1 equivalence class of vertices then there is a polygon with CSSI $P'$ with fewer sides than $P$ such that $S=P'/∼$.

Proof. Let $E$ be an equivalence class of vertices of $P$. Let $[v,w]=e_1$ be an edge with $v∈E,w∉E$. Let $e_2$ be the other edge adjacent to $v$.

Note that if $e_1∼e_2$ then $v∼w$. contradiction.

So $e_1$ gets identified to a edge $e_3$.

Continue decreasing $E$ till ${|E|}=1$. Then we get $P'$ with fewer edges. ◻

Definition. If the letters $x,x$ appear in a word we say $x,x$ is a standard pair. If $x,x^{-1}$ appear we say this is a reversed pair.

Proof. By induction on length $n$ of the word.

Say $n=4$.

Case 1.$x\,x^{-1}$ appears as subword

$x\,x^{-1}y\,y^{-1}=S^2$

$x\,x^{-1}y\,y=ℝP^2$

Case 2. $x\,x^{-1}$ does not appear as subword.

$x\,x\,y\,y=N_2$

After relabelling the start is $xy$

$xyxy=ℝP^2$

$xyx^{-1}y=xxyy=N_2$ (Lemma 1)

$xyxy^{-1}=xxy^{-1}y^{-1}=N_2$ (Lemma 2)

$xyx^{-1}y^{-1}=M_1$ ◻

Using Lemma 1 we can move all standard pairs at the beginning of the word. So we get a word $AB$, $A=x_1 x_1…x_k x_k$.

Definition. We say that the pairs $x,x^{-1}$ and $y,y^{-1}$ are interlocking if $y$ or $y^{-1}$ appears between $x,x^{-1}$ ie. $x…y…x^{-1}…y^{-1}$

Proof. By induction we prove that we can change $AB$ to $Ay_1 z_1 y_1^{-1}z_1^{-1}…y_n z_n y_n^{-1}z_n^{-1}$. Say at the $k$th step we have

$\displaystyle\underbrace{Ay_1 z_1 y_1^{-1}z_1^{-1}…y_k z_k y_k^{-1}z_k^{-1}}_E D\text{ and }x\text{ is first letter of }D$

Claim:$X,X^{-1}$ interlock ith some $y,y^{-1}$.

Suppose not. Then $v\nsim w$, so use Lemma 5 to get a smaller polygon.

So we have $ExFy{\color{red}{Gx^{-1}H}}y^{-1}k\xrightarrow{\text{Lemma 2}}Ex{\color{red}{FyHG}}x^{-1}y^{-1}k\xrightarrow{\text{Lemma 2},0}y^{-1}{\color{red}{kExHGF}}yx^{-1}$

Case 1. $n=0$. Then $S=N_k$.

Case 2. $k=0$. Then $S=M_h$.

Case 3. $k,n≠0$. By Lemma 3, $S=N_{k+2 n}$. ◻

Let $K$ be a finite simplicial complex with $n_i$ simplices of dimension $i$.

Define $χ(K)=\sum(-1)^i n_i$

Theorem. If $|K_1|$ homeomorphic to $|K_2|$, then $χ(K_1)=χ(K_2)$.

Easy to calculate $χ(M_g)=2-2 g,χ(N_h)=2-h$.

PREVIOUSTopologist's sine curve

NEXTTopology problem sheet 1

subsets of \(X/ℛ\)	⟷	saturated subsets of \(X\)
open subsets of \(X/ℛ\)	⟷	open saturated subsets of \(X\)
closed subsets of \(X/ℛ\)	⟷	closed saturated subsets of \(X\)