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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.

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If this is homework, you should add the "homework" tag to your question. Also, what have you tried so far? People will be more inclined to help you if you demonstrate that you have thought about this problem, rather than just trying to get someone to do it for you. –  Zev Chonoles May 3 '11 at 22:17
    
Do you know how to compute the fundamental group of a cell complex? (if yes, then use it:) For b): the glued point has a connected neighourhood which becomes disconnected when you remove the point. –  user8268 May 3 '11 at 22:26
    
Thanks, it's not from a homework, but from a past exam and I was just training. I posted mostly because I wanted to see how to write such a proof with van Kampen's theorem (just to figure out the amount of details required). –  TJIF May 3 '11 at 22:44
    
@user8268: Thanks, I think X is homeomorphic to the wedge sum of a torus and a sphere, I would however like to see a formal argument, if possible :-). –  TJIF May 3 '11 at 22:45
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Consider the torus with an added segment (outside it) that goes from x to y. If you contract this segment, you get your space but if you contract an arc within the torus that goes from x to y, you get the wedge sum of a torus and a circle. So your X is homotopy equivalent to (torus v circle) and the Seifert-van Kampen theorem gives you easily the fundamental group. –  PseudoNeo May 4 '11 at 0:09
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Here's how I might do it. (In fact, this is just the method suggested in PseudoNeo's comment above, but I wrote this out before I saw that particular comment, so I'm posting it for posterity.)

Consider the torus, and choose your two points $x$ and $y$. The space $X$ in which these two points are identified is homotopy equivalent to the space formed by attaching a copy of $I$ to the torus with endpoints $x$ and $y$, and since the torus is path-connected, this is homotopy equivalent to the wedge sum of a torus and a circle. By van Kampen's Theorem, and the homotopy invariance of $\pi_1$ we now know the fundamental group. Much like Jacob's proof, these homotopy equivalences are very easy to convince yourself of, but perhaps require a little more work to actually prove.

EDIT: My solution to part b) is obviously nonsense, homotopy equivalence doesn't preserve manifold-ness; consider a sphere with an interval attached at the basepoint, for instance. You can show that it's not homeomorphic to a closed compact surface by using van Kampen's theorem, the classification of such surfaces, the fundamental group calculated in part a), and the homeomorphism invariance of $\pi_1$. You can show that it's not even a 2-manifold by looking at Jacob's solution above.

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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 $H^\prime= H \circ (p \times id_{[0,1]}){}{}$.

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.

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@Beatty - see this question when I asked about the torus with $n$ points removed –  Juan S May 3 '11 at 23:13
    
Very nice ideas, although can you please elaborate more on why $S_{1} \wedge S_{1} \wedge S_{1}$ is a deformation retraction of $T - \left\{x,y\right\}$. I think it will shed some light on the general case. I see why we get $T^{2} \wedge S_{1}$ in the other solution, but I prefer your approach since it is more general. –  TJIF May 4 '11 at 23:40
    
Here ,two answers are given one says that $\pi_{1}(Y)$ is $\mathbb Z * \mathbb Z * \mathbb Z$ and other says $(\mathbb Z \times \mathbb Z)*\mathbb Z$. Strange! –  Shraddha Srivastava Jan 7 '13 at 12:56
    
@Jacob: You said V is simply connected so is $V \cap U$.I dont see it,could you please explain you argument? –  Shraddha Srivastava Jan 7 '13 at 12:58
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