1. 1. 1. Let $X$ and $Y$ be topological spaces. Provide two equivalent definitions for what it means for a function $f: X → Y$ to be continuous. You don't need to prove the equivalence between the two definitions.
      2. Provide two equivalent definitions for what it means for a subset $A ⊆ X$ of a topological space $X$ to be connected. You don't need to prove the equivalence between the two definitions.
      3. State what it means for a subset $A ⊆ X$ to be path-connected. Prove that every path-connected subspace is connected. [You may use without proof the fact that the interval $[0,1]$ is connected.]
      4. Prove that if $A ⊆ X$ is connected and $A ⊆ B ⊆ \bar{A}$, then $B$ is connected.
   2. Let $$ X=\left\{\left(\frac{\cos (x)}{1-\frac{1}{x}}, \frac{\sin (x)}{1-\frac{1}{x}}\right) ∈ ℝ^2: x>1\right\} ∪\left\{(x, y) ∈ ℝ^2: x^2+y^2=1\right\} $$ viewed as a subspace of $ℝ^2$.
      1. Is $X$ connected? Justify your answer carefully. [You may use without proof the fact that intervals are connected.]
      2. Prove that $X$ is not path connected. Justify your answer carefully.
2. 1. Let $X$ be a topological space, and let $∼$ be an equivalence relation on $X$. Denote by $X/\mmlToken{mi}∼$ the quotient set, and by \begin{aligned} π: X & → X/\mmlToken{mi}∼ \\ x & ↦[x] \end{aligned} the quotient map.
      1. Define the quotient topology on $X/\mmlToken{mi}∼$. Show that it is indeed a topology.
      2. Prove that the quotient topology is the finest topology (the topology with the most open sets) on $X/\mmlToken{mi}∼$ for which $π$ is continuous.
      3. What is the coarsest topology (the topology with the smallest number of open sets) on $X/\mmlToken{mi}∼$ for which the map $π$ is continuous? Justify your answer.
   2. 1. Define what it means for a topological space to be compact. Make sure to explain all the terms that you use. Prove that a set $X$ equipped with the discrete topology is compact if and only if it is finite.
      2. Prove that if $X$ is compact, then so is $X/\mmlToken{mi}∼$. If you use the result according to which the image of a compact space under a continuous map is compact, then also prove that result.
   3. 1. Now let $X$ be the unit interval $[0,1]$ equipped with its usual topology, and let $∼$ be the equivalence relation defined by $$ (x ∼ y) \stackrel{\text{def}}{⟺}(x=y  \text{ or }  (0<x<1\text{ and } x+y=1)) . $$ Show that the quotient space $X/\mmlToken{mi}∼$ is not Hausdorff. Justify your answer carefully.
      2. Which topologies on the two point set can one realise as quotient topologies of $[0,1]$ with respect to some equivalence relation? Justify your answer.
3. 1. 1. What is an abstract simplicial complex $K$ ?
      2. Provide an example of an abstract simplicial complex whose topological realisation is the two-dimensional torus $T^2$.
   2. 1. What is a closed combinatorial surface? Make sure to explain all the terms that you introduce.
      2. State the classification theorem for closed combinatorial surfaces. Once again, make sure to explain all the terms that you introduce.
      3. Let $P_1$ be a regular pentagon with edges $a, b, c, d, e$, listed in clockwise order, and oriented clockwise. Let $P_2$ be another pentagon with edges $α, β, γ, δ, ε$, again listed in clockwise order, and oriented clockwise: and Consider the surface $Σ$ obtained by taking the disjoint union of $P_1$ and $P_2$, and identifying $a$ to $α, b$ to $γ, c$ to $ε, d$ to $β$, and $e$ to $δ$. To which surface in the classification is $S$ homeomorphic? Justify your answer.
      4. Describe an explicit identification (homeomorphism) of the surface $Σ$ mentioned above with one of the surfaces listed in (b,ii).
   3. Show that there exists a continuous map $$ T^2 → K B $$ from the torus to the Klein Bottle, such that each point of the Klein Bottle admits exactly two preimages in $T^2$.

Solution

1. 1. 1. The function $f: X → Y$ is continuous if for every open set $U ⊆ Y$, its preimage $f^{-1}(U)$ is open in $X$. Equivalently, if for every closed set $V ⊆ Y$, its preimage $f^{-1}(V)$ is closed in $X$.
      2. The subspace $A$ is connected if its only non-empty clopen subset (in the subspace topology) is $A$ itself. Equivalently, $A$ is connected if there do not exist surjective continuous maps $A →\{*, *\}_{\text{discrete }}$.
      3. The space $A$ is path connected if $∀ a, b ∈ A$ there exists a continuous map $[0,1] → A$ which maps 0 to $a$, and 1 to $b$.
         If $A$ is not connected, then we may write it as a disjoint union of non-empty clopen subsets $A=V ⊔ W$. Pick $a ∈ V$ and $b ∈ W$, and a path $f:[0,1] → A$ which maps 0 to $a$, and 1 to $b$. The preimages $f^{-1}(V)$ and $f^{-1}(W)$ are clopen in $[0,1]$ and contain 0 and 1, respectively. This contradicts the connectedness of $[0,1]$.
      4. Assume that $A$ is connected and let $V ⊆ B$ be a non-empty clopen subset of $B$. We wish to show that $V=B$. Write $V$ as $V=U ∩ B$ for some $U ⊆ X$ open in $X$. Let $x ∈ V$ be point. Since $x ∈ \bar{A}$ and $U$ is open, the intersection $U ∩ A=V ∩ A$ is non-empty. Since $V$ is clopen in $B, V ∩ A$ is clopen in $A$, so we have $V ∩ A=A$ (by the connectedness assumption of $A$ ). In other words, $A ⊆ V$. Since $V$ is closed in $B$, we may write it as $V=B ∩ F$ for some $F ⊆ X$ closed in $X$. Since $A ⊆ F$ and $F$ is closed, we have $\bar{A} ⊆ F$. In particular $B ⊆ F$. So $V=B$.
   2. 1. Let $p:=(0,0) ∈ X$. The subspace $X':=\left\{\left(\frac{\cos (x)}{1-1 / x}, \frac{\sin (x)}{1-1 / x}\right)\right\}$ is homeomorphic to the half-open interval $[1, ∞)$, and thus connected. And the circle $X'':=\left\{(x, y): x^2+y^2=1\right\}$ is path-connected and hence connected. Therefore, any clopen subspace of $X$ must either contain $X'$ or be disjoint from $X'$, and the same for $X''$. The only subsets with that property are $∅, X, X'$, and $X''$. The latter two are not clopen: $X'$ is not closed, and $X''$ is not open. So the only clopen subset of of $X$ are $∅$ and $X$. In other words, $X$ is connected.
         
      2. If $X$ was path connected there would, in particular, exist a non-constant path $f:[0,1] → X$ with $f(0)=(0,0)$ and $f(1)=(1,0)$. Assume by contradiction that $f$ is such a path, and let $t=\sup \{x:{‖f(x)‖}<1\}$. By continuity, we have ${‖f(t)‖}=1$ (If $t=1$ this is obvious; otherwise this follows by the following argument: ${‖f(t)‖}=\left\|f\left(\lim _{n → ∞} t+\frac{1}{n}\right)\right\|=\lim _{n → ∞}\left\|f\left(t+\frac{1}{n}\right)\right\|=\lim _{n → ∞} 1=1$, using that the norm function is continuous). In particular, $t>0$ (because $f(0)=(0,0)$ and ${‖f(t)‖}=1$). Let $p:=f(t)$. Consider the open subset $U:=X∖\left\{-x p: x ∈ ℝ_{⩾ 0}\right\}$ of $X$, and note that it can be written as $U=\left(⋃_{n ∈ ℕ} U_n\right) ∪\left(X''∖\{-p\}\right)$, with $U_n=\left\{\left(\frac{\cos (x)}{1-1 / x}, \frac{\sin (x)}{1-1 / x}\right): x ∈\left(a_n, a_{n+1}\right)\right\} ⊂ X'$ for an appropriate sequence $\left\{a_n\right\}_{n ∈ ℕ}$. Each $U_n$ is clopen in $U$, as it can be written as both $U_n=U ∩\left\{\left(y \frac{\cos (x)}{1-1 / x}, y \frac{\sin (x)}{1-1 / x}\right): x ∈\left(a_n, a_{n+1}\right), y ∈(1-ϵ, 1+ϵ)\right\}$ for suitably small $ϵ>0$, and $U_n=U ∩\left\{\left(\frac{\cos (x)}{1-1 / x}, \frac{\sin (x)}{1-1 / x}\right): x ∈\left[a_n, a_{n+1}\right]\right\}$. Since $U$ is open in $X$ and $f(t)=p ∈ U$, by continuity, $∃ ϵ>0$ such that $f((t-ϵ, t]) ⊂ U$. Let $y ∈(t-ϵ, t]$ be such that ${‖f(y)‖}<1$. Then $f(y)$ lies in one of the clopen subset $U_n$. Consider the preimage of $U_n$ under the map $f:(t-ϵ, t] → U$. It is clopen in $(t-ϵ, t]$ because $f$ is continuous. It is non-empty because $f(y) ∈ U_n$. It is not the whole of $(t-ϵ, t]$ because $f(t)=p$. This contradicts the connectedness of the interval $(t-ϵ, t]$. We have reached a contradiction. The points $(0,0)$ and $(1,0)$ cannot be connected by a continuous path, and $X$ is not path connected.
2. 1. 1. A subset $U ⊆ X/\mmlToken{mi}∼$ is open in the quotient topology if its preimage $π^{-1}(U)$ is open in $X$. We check that $∅$ and $X/\mmlToken{mi}∼$ are open in the quotient topology: indeed $π^{-1}(∅)=∅$ and $π^{-1}(X/\mmlToken{mi}∼)=X$ are open in $X$. We check that the intersection of two opens $U, V ⊆ X/\mmlToken{mi}∼$ is again open: indeed $π^{-1}(U ∩ V)=π^{-1}(U) ∩ π^{-1}(V)$ is open in $X$. We check that the union of a family of opens $U_i ⊆ X/\mmlToken{mi}∼$ is again open: indeed $π^{-1}\left(⋃ U_i\right)=⋃ π^{-1}\left(U_i\right)$ is open in $X$.
      2. Let $𝒯_\text{quot}$ be the quotient topology $X/\mmlToken{mi}∼$, and let $𝒯$ be a topology on $X/\mmlToken{mi}∼$ with respect to which the map $π$ is continuous. Then, for every open $U ⊂ X$, its preimage $π^{-1}(U)$ is open in $𝒯$. So the quotient topology is finer (or equal) than the topology $𝒯$. That is, $𝒯 ⊆ 𝒯_\text{quot}$. The quotient topology $𝒯_\text{quot}$ makes $π$ continuous. So $𝒯_\text{quot}$ is the finest topology which makes $π$ continuous.
      3. The coarsest topology on $X/\mmlToken{mi}∼$ is the indiscrete topology: the topology for which only the empty set and the whole set are open. Any map to a space equipped with the indiscrete topology is continuous, in particular $π: X →(X/\mmlToken{mi}∼)_{\text{indiscrete }}$ is continuous. Any other topology on $X/\mmlToken{mi}∼$ has more open sets (whether or not it makes $π$ continuous).
   2. 1. A topological space $X$ is said to be compact if every open cover (collection of open subsets whose union is $X$) admits a finite subcover (sub-collection whose union is still $X$). If the space $X$ is finite, then any cover is finite as there are only finitely many subsets of $X$. So there is nothing to prove. If the space $X$ is infinite, then the collection of all singletons $\{x\}$ (indexed by the points of $X$) is an open cover which does not admit a finite subcover.
      2. Let $X$ be compact and let $\left\{U_i\right\}_{i ⊆ I}$ be an open cover of $X/\mmlToken{mi}∼$. Then $\left\{π^{-1}\left(U_i\right)\right\}_{i ∈ I}$ is an open cover of $X$. Since $X$ is compact, there exists a finite subcover $\left\{π^{-1}\left(U_i\right)\right\}_{i ⊂ I'}$ (for some finite subset $I' ⊆ I$ ). Then $\left\{U_i\right\}_{i ⊂ I'}$ is the desired subcover of $X/\mmlToken{mi}∼$.
   3. 1. The quotient is the interval $[0,1 / 2]$ with the points 0 "doubled". Let us call $0_1 ∈ X/\mmlToken{mi}∼$ the image of $0 ∈[0,1]$, and $0_2 ∈ X/\mmlToken{mi}∼$ the image of $1 ∈[0,1]$. If $U$ is a neighbourhood of $0_1$, then $π^{-1}(U)$ is a neighbourhood of 0 and thus contains an interval $\left[0, ϵ_1\right)$. So $U$ contains $\left(0, ϵ_1\right) ⊂ X/\mmlToken{mi}∼$. If $V$ is a neighbourhood of $0_2$, then $π^{-1}(V)$ is a neighbourhood of 1 and thus contains an interval $\left(1-ϵ_2, 1\right]$. So $V$ contains $\left(0, ϵ_2\right) ⊂ X/\mmlToken{mi}∼$. It follows that $U ∩ V$ contains $\left(0, \min \left\{ϵ_1, ϵ_2\right\}\right) ⊂ X/\mmlToken{mi}∼$, so $U ∩ V ≠ ∅$.
      2. There are exactly three topological spaces (up to homeomorphism) whose underlying set has two points: the set $\{0,1\}$ with the discrete topology; the set $\{0,1\}$ with the indiscrete topology; and the set $\{0,1\}$ with the Sierpinski topology. The first one is not connected and so cannot be the image of a connected space (such as $[0,1]$) under a continuous map. So $\{0,1\}$ with the discrete topology cannot the written as a quotient of $[0,1]$. The two other topologies can be obtained as quotients of $[0,1]$.
         For the indiscrete topology, neither of $\{0\},\{1\}$ is open, one can pick e.g. the equivalence relation on $[0,1]$ whose two equivalence classes are $ℚ ∩[0,1]$ and $(ℝ∖ ℚ) ∩[0,1]$.
         For the Sierpinski topology, only one of $\{0\},\{1\}$ is open, one can take e.g. the equivalence relation on $[0,1]$ whose two equivalence classes are $\{0\}$ and $(0,1]$.
3. 1. 1. An abstract simplicial complex is a set $K$ together with a collection of non-empty finite subsets $F ⊆ 𝒫(K)∖ ∅$ (called simplices of $K$) such that every singleton is in $F$, and $\left(f ∈ F\right.$ and $\left.f' ⊂ f\right) ⇒ f' ∈ F$.
      2. Consider the following triangulation of the torus: where the vertices of the triangulation are named $a, b, c, d, e, f, g, h, i$. The desired abstract simplicial complex has underlying set $K=\{a, b, c, d, e, f, g, h, i\}$ and set of simplices consisting of all the singletons $\{a\},\{b\}, …,\{i\}$, the 27 2-element subsets $\{a, b\},\{a, c\},\{a, d\}, …$ that correspond to edges in the above picture, and the 18 3-element subsets $\{a, b, d\},\{a, b, h\}, …$ which appear as triangles in the above picture.
   2. 1. A closed combinatorial surface is a connected finite simplicial complex $K$ (alternatively, the geometric realisation of a finite simplicial complex $K$ ) such that for every vertex $v ∈ K$ the link of $v$ is a simplicial circle. Here, the link $\DeclareMathOperator{\lk}{lk}\lk(v)$ of a vertex $v ∈ K$ is the set of $\lk(v)=\{w ∈ K$ : $\{v, w\}$ is a simplex of $K\}$, where a subset $f ⊂ \lk(v)$ is declared to be a simplex of $\lk(v)$ if $f ∪\{v\}$ is a simplex of $K$. Finally, a simplicial circle is a simplicial complex whose topological realisation is $S^1$.
      2. The classification theorem says that (the topological realisation of) any closed combinatorial surface is homeomorphic to either the 2-sphere $S^2$, or $S^2$ with $n$ handles attached for $n ⩾ 1$, or $S^2$ with $n$ crosscaps attached for $n ⩾ 1$, and that these surfaces are all pairwise non-homeomorphic. Here, $S^2$ with $n$ handles attached is the quotient of a $(4 n)$-gon by identifying its sides according to the word $x_1 y_1 x_1^{-1} y_1^{-1} x_2 y_2 x_2^{-1} y_2^{-1} … x_n y_n x_n^{-1} y_n^{-1}$, and $S^2$ with $n$ crosscaps attached is the quotient of a $(2 n)$-gon by identifying its sides according to the word $x_1 x_1 x_2 x_2 x_3 x_3 … x_n x_n$.
      3. One possible answer is to observe that $S$ is orientable with Euler characteristic $χ(S)=1-5+2=-2$ (this requires noting that all the vertices of $P_1$ and of $P_2$ get identified in $S$ ). It is therefore homeomorphic to a sphere with two handles by the classification theorem.
      4. Glue $P_1$ and $P_2$ along $a ∼ α$ to get an 8-gon whose sides are identified according to the word $b c d e c^{-1} e^{-1} b^{-1} d^{-1}$. Cutting along the dotted line, and re-gluing the cyan triangle as indicated yields an 8-gon whose sides are identified according to the word $c x c^{-1} x^{-1} d b^{-1}$ Upon replacing $b$ by $b^{-1}$, this is exactly the presentation of a sphere with two handles.
   3. Let us write $T^2=$ and $K B=$ as quotients of $[0,1]^2$ by the usual identifications of opposite edges. Then the map $[0,1]^2 →[0,1]^2$ given by $(x, y) ↦(x, 2 y)$ if $y ⩽ 1 / 2$ and $(x, y) ↦(1-x, 2 y-1)$ if $y ⩾ 1 / 2$ descends to a map $T^2 → K B$ with all the desired properties.