Homology of the two-dimensional torus with $n$ points removed

Question

(I know there already exists a question addressing this: Homology of a torus with points removed but I’m trying to ask questions about the precise calculations here, which weren’t addressed in that question.)

I’m trying to solve the following exercise: Let $A$ be a finite set of points in the two-dimensional torus $T$. Compute the homology groups of $T \setminus A$ and $T /A$.

My attempt so far:

I think I can use the excision theorem with $A \subset U \subset T$ where $U$ is an open set on the torus that contains all the points. I.e. $U\setminus A$ is homeomorpic to a disc with $|A|$ points removed. Thus by excision: $H_n(T\setminus A, U \setminus A) \cong H_n(T,U)$. Now we can consider the LES:

$$ \dots \rightarrow H_{n+1}(T,U) \rightarrow \tilde{H}_n(U) \rightarrow \tilde{H}_n(T) \rightarrow H_n(T,U) \rightarrow \tilde{H}_{n-1}(U) \rightarrow \dots $$

We also know that: $$ \tilde{H}_n(T)=\begin{cases} 0, \ n=0 \\ \mathbb{Z}^2, \ n=1 \\ \mathbb{Z}, \ n=2 \\ 0, \ n\geq 3 \end{cases}, \quad \tilde{H}_n(U)=0 $$

Thus I think I can deduce (by considering the LES):

$$H_n(T,U)=\begin{cases} 0, \ n=0 \\ \mathbb{Z}^2, \ n=1 \\ \mathbb{Z}, \ n=2 \\ 0, \ n\geq 3 \end{cases} $$

Now since there is a homeomorphism of pairs $(T\setminus A, U \setminus A) \rightarrow (T/A\setminus A/A, U/A \setminus A/A) $ and again using excision: $$ H_n(T,U) \cong H_n(T\setminus A, U \setminus A) \cong H_n(T/A, U/A) $$

So we can finally consider the LES:

$$ \dots \rightarrow H_{n+1}(T \setminus A,U \setminus A) \rightarrow \tilde{H}_n(U \setminus A) \rightarrow \tilde{H}_n(T \setminus A) \rightarrow H_n(T \setminus A,U \setminus A) \rightarrow \tilde{H}_{n-1}(U \setminus A) \rightarrow \dots $$

Because we know the Homology of $S^1$ and $U \setminus A \simeq S^1 \vee \dots \vee S^1$ ($|A|$ times) we get that (by Mayer-Vietoris): $$ \tilde{H}_n(U \setminus A)=\begin{cases} 0, \ n=0 \\ \mathbb{Z}^{|A|}, \ n=1 \\ 0, \ n\geq 2 \end{cases} $$

So now I have all the information for my LES: $n=1$: $$ \mathbb{Z} \rightarrow \mathbb{Z}^{|A|} \rightarrow \tilde{H}_1(T \setminus A) \rightarrow \mathbb{Z}^2 \rightarrow 0 $$

$n=2$: $$ 0 \rightarrow 0 \rightarrow \tilde{H}_2(T \setminus A) \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}^{|A|} $$

But I can’t seem to find a precise argument now to deduce the groups $\tilde{H}_n(T\setminus A)$ from this. Also to then find $\tilde{H}_n(T/A)$ it seems to me that I need $\tilde{H}_n(U/A)$, could someone maybe help me how to deduce them?

Answer 1

In order to proceed with your approach, we need to determine what the boundary maps do. That means we need to track what happens to a generator of the torus homology under $H_\ast(T)\cong H_\ast(T,U)\cong H_\ast(T\setminus A,U\setminus A)$. At first glance that's not very pleasant, but you can see from the square-quotient description, $U$ being some open subsquare and all the points of $A$ lying in $U$, that it will send the generator - which is just a standard homeomorphism $\Delta^2\cong I^2$ followed by the quotient $I^2\twoheadrightarrow T$ - to a subdivided chain whose boundary is a encircles the circumference of $U$, circumnavigating the points of $A$ with winding one, and by understanding how $U\setminus A\simeq\bigvee_AS^1$ works you realise the overall map $H_2(T)\cong H_2(T\setminus A,U\setminus A)\to H_1(U\setminus A)$ corresponds to (up to sign) $1\mapsto(1,1,1,\cdots,1),\,\Bbb Z\to\Bbb Z^{|A|}$. Then $H_2(T\setminus A)=0$ follows, as this injects, and $H_1(T\setminus A)\cong\Bbb Z^2\oplus\Bbb Z^{|A|}/(1,1,\cdots,1)\Bbb Z\cong\Bbb Z^{2+|A|-1}=\Bbb Z^{|A|+1}$ as expected. You get that because $\Bbb Z^2$ is free, so the short exact sequence splits.

You may object that that's not especially rigorous. Fine. We can make it more precise with a Mayer-Vietoris sequence. Use the square model again, and decompose the torus as the union of $X$ which is the boundary padded with some (half-)open rectangular strips avoiding $A$, and where $Y=U$ is what you get when you remove $A$ from an open rectangle containing all points of $A$ and just overlapping with $X$ in another open rectangle, but not touching the boundary. When you apply the quotient map, $T\setminus A=X\sqcup Y$ still holds as an open decomposition. So M-V applies. Notice $X\cap Y\simeq S^1$ and that the generator of its first homology does indeed, by explicit construction of $Y\simeq\bigvee_AS^1$, correspond to $(1,1,\cdots,1)$. Moreover $X$ deformation retracts onto the image of the boundary of the square, which is just $S^1\vee S^1$, and the generator of $H_1(X\cap Y)$ corresponds to $(1-1,1-1)=(0,0)$ in $\Bbb Z^2\cong H_1(S^1\vee S^1)\cong H_1(X)$. This is rigorous because we explicitly know the generators in all cases. Then you get an M-V sequence isomorphic to $0\cdots0\to H_2(T\setminus A)\to\Bbb Z\overset{\langle-\Delta,0\rangle}{\to}\Bbb Z^{|A|}\oplus\Bbb Z^2\to H_1(T\setminus A)\to0$ which gives the same results.