What is this quotient space of the torus?

Quotient space of the torus with two points identified

• Take the torus $T=S_1 \times S_1$. Choose two points $x,y \in T$ and define a quotient topology by identifying $x$ and $y$. Let $X$ denote the quotient space. Prove that: a) Compute the fundamental group of $X$. b) Prove that $X$ is not homeomorphic to a surface.

• Answer:

  So for part a. Let $T:=S_1 \times S_1$ be the torus. Suppose that $x,y \in T$ such that $x \neq y$ and define an equivalence relation $\sim$ that identifies $x,y$ and nothing else. Then let $Y:= X / \sim$ and $p: T \rightarrow Y$ be the quotient map. Note that $p|_{T \setminus \{x,y\}}$ is injective and so homeomorphic on its image. Then for SVK we may pick U to be $p(T \setminus\{x,y\})$ and $V$ to be the image of some regular euclidean ball in $T$ not containing $x$ or $y$. From this we can deduce that $\pi_1(Y) \cong \pi_1(T \setminus \{x,y\})$ since $V$ is simply connected and so is $V \cap U$. Then to determine the fundamental group of $T \setminus \{x,y\}$, draw a square with the torus' identifications and remove two points. It shouldn't be hard to convince yourself that $S_1 \vee S_1 \vee S_1$ is a deformation retraction of this space. It's a bit more work to prove it. The idea is to create a deformation on the square that respects the identifications of $p$ so it induces the desired deformation on the quotient space. For part b) as user user8268 said if you look at a "small" enough open set around $p(y)$ then removing $p(y)$ will make it disconnected, which in turn implies it cannot be locally homeomorphic to $\mathbb R^2$ everywhere. Edit Update: Torus with n holes removed. Claim: The torus with n holes removed has fundamental group of the wedge of $n+1$ circles. The basic idea here is that we can work purely in the square and as long as we don't mess with the "edges" of the square then we can get the desired homotopy. Here is a useful fact: If $p: X \rightarrow Y$ is a quotient map and $H: X \times I \rightarrow Z$ is a homotopy that respects the identifications of $p$, that is if for $x, y \in X$ $p(x)=p(y)$ then $h(x,t)=h(y,t)$ for all time $t$. Then there is a natural homotopy $H^\prime: Y \times I \rightarrow Z$ such that $H^\p= H \circ (p \times id_{[0,1]})$. This allows me to construct a homotopy on the square with $n$ points removed and then use it to induce a homotopy on the torus. This lemma isn't too hard to prove, but you need to know that a quotient map crossed with the identity of a locally compact space is still a quotient map, which is very challenging to prove. Lee proves it his book on topological manifolds, it used to be at the end of the chapter on compactness, but I'm not sure anymore since he added paracompactness in the second edition. Anyway, let $a_1,\dots,a_n \in \mathbb I^2$ where $\mathbb I=[0,1]$. Assume that $a_1,\dots,a_n$ are not boundary points, for our purposes this is fine. I'd like to draw a picture now, but I can't. Take the $n$ points and move them such that $a_i=(i/n,1/2)$ so we're arranging them along the center strip each at distance $n$ away. Hopefully you're familiar with how to retract a punctured disk onto its boundary. Well we're going to do that for each one of these points except we'll be adding a vertical bar in between each of the points. Now we can create a homotopy that retracts onto these vertical bars, so that $H: \mathbb I^2 \times \mathbb I \rightarrow \mathbb I$ for $H(x,0)=id_{\mathbb I^2}$ but for $H(\mathbb I^2 \times \{1\})=X$ where $$X:= \mathsf{bd}(I^2) \cup \bigcup_{i=1}^{n-1} \{\frac{2i+1}{2n}\} \times \mathbb I$$, that is we get just the boundary of the square and these vertical bars connecting the top and the bottom. Note that if we compose this map with the quotient map from the square to the n-holed torus we get a homotopy that respects the identifications of the torus, so we can apply the previous lemma and get the desired homotopy. Finally it isn't hard to see that the image of $X$ under our quotient map is the wedge of $n+1$ circles. You can use SVK to prove this without too much work. Disclaimer: I've left a lot out, because it's a bit of a tedious proof and it's sort of hard to get a good deal of it across without drawing a few pictures.