Topology paper 2022

1. 1. Define the operation of connected sum of surfaces.

      The connected sum of $X$ and $Y$ is the surface$$X \# Y:=(X \backslash\mathring D) ⊔\left(Y \backslash\mathring{D'}\right) / ∼$$ where $D ⊂X$ and $D' ⊂Y$ are embedded discs, and the equivalence relation ∼ identifies $∂ D$ and $∂ D'$ via some homeomorphism $φ: ∂ D → ∂ D'$ [Not required: If $X$ and $Y$ are surfaces with boundary, then $D$ and $D'$ should not meet the boundary].

   2. State the classification theorem of surfaces in terms of connected sums of known surfaces.

      Every closed surface is homeomorphic to either $S^2$, a connected sum of copies of the torus $T^2$, or a connected sum of copies of the projective plane $ℝP^2$. Moreover, the surfaces in the above list are pairwise non-homeomorphic.

2. Prove that these two spaces are homeomorphic: (the connected sum of two Möbius bands, and the Klein bottle with two discs removed). [Hint: describe (and prove) how the Möbius band is related to $ℝP^2$, then describe (and prove) how two copies of $ℝP^2$ are related to the Klein bottle.]

   The Möbius band is homeomorphic to $ℝP^2$ with one disc removed: Lemma 5.12
   To see this, consider the following construction of $ℝP^2$: and remove this disc:
   So the connected sum of two Möbius bands is homeomorphic to $ℝP^2 \# ℝP^2$ with two discs removed.
   It remains to show that $ℝP^2 \# ℝP^2$ is homeomorphic to a Klein bottle: Example 5.22 The boundary words xy⁻¹xy and xxyy represent homeomorphic surfaces.

3. Let $S^1 ⊂ℝ^2$ be the unit circle.
   1. Is ℝ homeomorphic to a subspace of $S^1$ ? If yes:
      α) is it homeomorphic to an open subspace of $S^1$ ?

      Yes: $S^1$ minus a point is homeomorphic to an open interval. The latter is homeomorphic to ℝ.

      β) is it homeomorphic to a closed subspace of $S^1$ ?

      No: Any closed subset of $S^1$ is compact, but ℝ is not compact.

   2. Is $S^1$ homeomorphic to a subspace of ℝ ?
      If yes: α) is it homeomorphic to an open subspace of ℝ ? β) is it homeomorphic to a closed subspace of ℝ ?

      No: Any subset of ℝ (with at least three points) has the property that it becomes disconnected after removing some point. But $S^1$ is connected and remains connected after removing any point.

4. 1. Prove that $U ⊂ℝ$ is open in ℝ if and only if $U ∩[a, b]$ is open in $[a, b]$ for all $a, b∈ ℝ$.

      If $U$ is open, then $U ∩[a, b]$ is open in $[a, b]$ by definition of the subspace topology. Assume that $U ∩[a, b]$ is open in $[a, b]$ for all $a, b∈ ℝ$. We need to show that for every point $x∈ U$ there exists $ε>0$ such that $(x-ε, x+ε) ⊂U$. Pick any interval $[a, b]$ containing $x$ in its interior. Since $U ∩[a, b]$ is open in $[a, b]$, there exists $ε>0$ such that $(x-ε, x+ε) ⊂U ∩[a, b]$. In particular, $(x-ε, x+ε) ⊂U$.

   2. Let $X$ be a topological space. Is it true that a subset $U ⊂X$ is open in $X$ if and only if $U ∩ V$ is open in $V$ for all closed subsets $V ⊊ X$ ?
      α) Prove that the answer is "yes" when $X$ is Hausdorff.

      (If $U$ is open, then $U ∩ V$ is open in $V$ by definition of the subspace topology.) Assume $U ∩ V$ is open in $V$ for all closed subsets $V ⊊ X$. We'll show that $U$ is open by proving that every point $x∈ U$ admits an open neighbourhood contained in $U$. Pick $y∈ X$ distinct from $x$ (if $X$ has only one point, then $U$ is trivially open). Since $X$ is Hausdorff, there exist disjoint open subsets $A, B ⊂X$ with $x∈ A$ and $y∈ B$. Let $V:=X \backslash B$, so that$$x∈ A ⊂V ⊊ X$$By assumption, $U ∩ V$ is open in $V$. Therefore $U ∩ A$ is open in $A$. Since $A$ is open in $X, U ∩ A$ is also open in $X$. $U ∩ A$ is the desired neighbourhood of $x$.

      β) Prove that the answer is "no" in general.

      Let $X$ be the two point space equipped with the indiscrete topology, and let $U$ be a subspace consisting of just one of the two points. $U$ is not open, but $U ∩ ∅$ is open in $∅$ (and $∅$ is the only closed subset of $X$ which is not the whole space).

1. 1. Define what it means for a topological space to be compact. [If you use words such as "cover" and "subcover", you must also define these words.]

      A topological space $X$ is compact if every open cover $\left\{U_i\right\}_{i∈ I}$ (an open cover is a collection of open sets whose union is $X$) admits a subcover $\left\{U_i\right\}_{i∈ J}$ (a subcover is a sub-collection whose union is still $X$) where $J ⊂I$ is finite.

   2. A topological space $X$ is called locally compact if for every point $x∈ X$ there exists $U ⊂X$ open and $K ⊂X$ compact such that $x∈ U ⊂K$.
      α) Prove that ℝ with its usual topology is locally compact.

      Given $x∈ ℝ$, we have $x∈(x-1, x+1) ⊂[x-1, x+1]$. In other words, $[x-1, x+1]$ is a compact neighbourhood of $x$.

      β) Is ℚ (equipped with the subspace topology from ℝ) locally compact? Justify.

      No point $x∈ℚ$ admits a compact neighbourhood. If $V ⊂ℚ$ is a neighbourhood of $x$, then $[x-ε, x+ε]∩ℚ⊂V$ for some $ε>0$. Since $[x-ε, x+ε]∩ℚ$ is not complete, it is not compact, but $[x-ε, x+ε]∩ℚ$ is closed in $V$, so $V$ is not compact.

2. Let $D^2 ⊂ℝ^2$ be the closed unit disc.
   1. Prove that any bounded convex polygon $V ⊂ℝ^2$ is homeomorphic to $D^2$.

      We may assume without loss of generality (by performing a translation) that $V$ contains 0 in its interior. Let $ρ_θ∈ ℝ⩾ 0$ be defined by the requirement that $\left[0, ρ_θ ⋅ θ\right]=V ∩\left(ℝ_{⩾ 0} ⋅ θ\right)$. Note that this is well-defined: since $V$ is convex, the intersection $V ∩\left(ℝ_{⩾ 0} ⋅ θ\right)$ is also convex, in particular connected. The function $θ ↦ ρ_θ$ is clearly continuous (it is piecewise algebraic). The map $D^2 → V$ given, in polar coordinates, by $(r, θ) ↦\left(ρ_θ r, θ\right)$ is the desired homeomorphism. It is continuous at non-zero points because $ρ_θ$ is continuous. It is continuous at zero because $ρ_θ$ is bounded. Finally, it maps bijectively $D^2$ onto $V$, and is thus a homeomorphism (a continuous bijection between compact Hausdorff spaces is a homeomorphism).

   2. Consider the equivalence relation $∼$ on the unit square $[0,1]^2$ given by $(x, y) ∼\left(x', y'\right)$ if $(x, y)=\left(x', y'\right)$ or if $x=x'=0$. Prove that the quotient space $[0,1]^2 / ∼$ is homeomorphic to $D^2$.

      Let $T ⊂ℝ^2$ be the triangle with vertices $(0,0),(1,0)$ and $(1,1)$. The map $f:[0,1]^2 →T$ given by $f(x, y)=(x, x y)$ is continuous and surjective. It satisfies $f(x, y)=f\left(x', y'\right)$ iff $(x, y) ∼\left(x', y'\right)$ therefore it descends to a continuous map (indeed a continuous bijection) $\bar{f}:[0,1]^2 /\mmlToken{mi}∼ → T$. The $\operatorname{map} \bar{f}$ is a homeomorphism (because a continuous bijection from a compact space to a Hausdorff space is a homeomorphism: $[0,1]^2/\mmlToken{mi}∼$ is compact because it's the image of $[0,1]^2$ under a continuous map; $T$ is Hausdorff because it's a subspace of $ℝ^2$ ). Finally, $T$ is homeomorphic to $D^2$ by (b.i).

3. Let $D^2 ⊂ℝ^2$ be the closed unit disc, and let $H=ℝ× ℝ_{⩾ 0} ⊂ℝ^2$ be the closed upper half plane. Consider the following four topological spaces:

   $X_1:=D^2 \backslash\{(1,0)\}$	$X_2:=\stackrel{∘}{D}^2 ∪\left(D^2 ∩ H\right)$	$X_3:=\stackrel{∘}{D}^2 ∪\left(D^2 ∩ \stackrel{∘}{H}\right)$	$X_4:=\stackrel{∘}{D}^2 ∪\{(1,0)\}$

   all of them equipped with the subspace topology from $ℝ^2$.
   1. Which two of the above four spaces are homeomorphic? Prove that they are indeed homeomorphic. [Hint: Use (b) (ii) to construct the desired homeomorphism.]

      Let $\bar{f}:[0,1]^2 /\mmlToken{mi}∼ → T$ be the homeomorphism constructed in part (b), and let $p∈[0,1]^2 /\mmlToken{mi}∼$ be the point corresponding to the subset $A:=\{0\} ×[0,1] ⊂[0,1]^2$ (note that $A$ is an equivalence class for the equivalence relation ∼). By (b.i), we have the following two homeomorphisms: $X_1≅T \backslash\{(0,0)\}$ and $X_3≅[0,1]^2 \backslash A$. The map $\bar{f}$ induces a homeomorphism $\left([0,1]^2 /\mmlToken{mi}∼\right) \backslash\{p\}≅T \backslash\{(0,0)\}$. The composite map $[0,1]^2 \backslash A ↪[0,1]^2↠[0,1]^2 /\mmlToken{mi}∼$ induces a homeomorphism $[0,1]^2 \backslash A≅([0,1]^2 /\mmlToken{mi}∼)\backslash\{p\}$. Assembling all these maps, we get a homeomorphism$$X_3 \stackrel{≅}{⟶}[0,1]^2 \backslash A \stackrel{≅}{⟶}\left([0,1]^2 /\mmlToken{mi}∼\right) \backslash\{p\} \stackrel{≅}{⟶} T \backslash\{(0,0)\} \stackrel{≅}{⟶} X_1 .$$

   2. Prove that no other two of the above four spaces are homeomorphic. [Hint: Which ones are locally compact?]

      We count the number of points that do not admit compact neighbourhoods: $X_1$ and $X_3$ admit no such points (they are locally compact). $X_2$ has two such points. $X_4$ has one such point.

1. Provide an example of an abstract simplicial complex whose topological realisation is homeomorphic to ℝ.

   Take $ℤ$ as the vertex set, and $\{\{n\}: n∈ ℤ\} ∪\{\{n, n+1\}: n∈ ℤ\}$ as the set of simplices.

2. 1. Let $X$ be a topological space. Let $Q$ be a set, and let $f: X ↠ Q$ be a surjective function.
      α) The map $f$ induces a bijection $\bar{f}: X / ∼ → Q$, for a certain equivalence relation ∼ on $X$. Describe the equivalence relation ∼.

      The equivalence relation ∼ is given by declaring $x ∼ y$ when $f(x)=f(y)$.

      β) Describe the unique topology on $Q$ which makes $\bar{f}$ into a homeomorphism (where $X / ∼$ is equipped with the quotient topology).

      A subspace $U ⊂Q$ is open in this topology iff $f^{-1}(U)$ is open in $X$.

   2. Define what it means for a continuous map $f: X → Y$ between topological spaces to be a quotient map.

      $f$ is a quotient map if $\left\{\begin{aligned}&f\text{ is surjective}\\&U\text{ is open in $Y$ iff $f^{-1}(U)$ is open in $X$}\end{aligned}\right.$

   3. Let $S^1$ be the set of complex numbers of norm one, topologised as a subset of $ℂ$. Fix $b ⩾ 1$ a real number, and consider the half-open interval $[0, b) ⊂ℝ$. Is the map \begin{aligned} f:[0, b) & ⟶ S^1 \\ x & ↦ e^{2 π i x} \end{aligned} a quotient map? Justify your answer. [If the answer depends on the choice of $b$, treat the different cases separately.]
      • This is not true when $b=1$ as, in that case, the map $f:[0,1) → S^1$ is a bijection. A bijective quotient map is a homeomorphism. But $S^1$ is compact whereas $[0,1)$ is not. So $f$ is not a quotient map.
      • This is true when $b>1$. Given a subset $U ⊂S^1$ whose preimage is open in $[0, b)$, we must show that $U$ is open in $S^1$. Pick a point $z∈ U$. We will show that there exists an open arc containing $z$, entirely contained inside $U$. Let $x$ be the unique preimage of $z$ under $f$ such that $x∈(0,1]$. Since $x$ lies in the interior of $[0, b)$ and $f^{-1}(U)$ is open in $[0, b)$, there exists $ε>0$ such that $(x-ε, x+ε) ⊂f^{-1}(U)$. It follows that $f((x-ε, x+ε)) ⊂U$.

Declare $x≡ x'$ if their images in $Z$ are equal. The composite $X → Y → Z$ induces a continuous bijection$$f: X /\mmlToken{mi}≡ → Z$$that fits into a commutative triangle A subspace $U ⊂Z$ is open iff its preimage in $Y$ is open (by definition of the quotient topology on $Y /\mmlToken{mi}≈$) iff its preimage in $X$ is open (by definition of the quotient topology on $X /\mmlToken{mi}∼)$. Similarly, a subspace $U' ⊂X /\mmlToken{mi}≡$ is open iff its preimage in $X$ is open. The bijective map $f$ induces a bijection between the open subsets of $X /\mmlToken{mi}≡$ and the open subsets of $Z$. So $f$ is a homeomorphism.

The first space $S^2 / ∼$ is the standard definition of $ℝP^2$. The second space can be viewed as an iterated quotient$$\left(S^2 /(x, y, z) ∼(x, y,-z)\right) /(x, y, 0) ∼(-x,-y, 0)$$as in (c). The first quotient $S^2 /(x, y, z) ∼(x, y,-z)$ is homeomorphic to $D^2$ via the quotient map $S^2 → D^2:(x, y, z)↦(x, y)$ (a bijective continuous map between compact Hausdorff spaces is a homeomorphism). The equivalence relation $(x, y, 0) ∼(-x,-y, 0)$ on points of $S^2$ induces the equivalence relation $≈$ on $D^2$ that identifies antipodal points of the boundary. The quotient of $D^2 / ≈$ is another standard definition of $ℝP^2$. Explicitly, the embedding $D^2↪S^2:(x, y) ↦\left(x, y, \sqrt{1-x^2-y^2}\right)$ induces a homeomorphism $S^2 / ∼ → D^2 / ≈$ (a bijective continuous map between compact Hausdorff spaces is a homeomorphism).
$S^2 /(x, y, z) ∼(-x,-y,z)$ is isomorphic to the Klein bottle $N_2$.