Homology of torus *without* cellular boundary formula

Question

Decomposing a torus $T$ as $\{e^2,e^1_1,e^1_2,e^0\}$, where $e^i$ has dimension $i$, the problem is to compute $H_*(T)$. I haven't learned the cellular boundary formula yet. The hint is to note that the $1$-skeleton is $S^1\vee S^1$, so that $H_2(T,S^1\vee S^1)=\mathbb Z$ and $H_1(T,S^1\vee S^1)=0$.

I tried using the long exact sequence for the pair $(T,S^1\vee S^1)$ to solve this. In particular, since $H_2(S^1\vee S^1)=0$ and $H_1(S^1\vee S^1)=\mathbb Z\oplus\mathbb Z$, I get the exact sequence $$0\to H_2(T)\to\mathbb Z\to\mathbb Z\oplus\mathbb Z\to H_1(T)\to0.$$ The only problem is that I don't know what the map $\mathbb Z\to\mathbb Z\oplus\mathbb Z$ is. If it is an injection, then I get the (incorrect) solution that $H_2(T)=0$ and $H_1(T)=\mathbb Z$, while if $\mathbb Z\to\mathbb Z\oplus\mathbb Z$ is the zero map, then I get the (correct) solution that $H_2(T)=\mathbb Z$ and $H_1(T)=\mathbb Z\oplus\mathbb Z$. I don't think there's any contradiction in either case, so it seems like this won't be helpful.

The other option is to use cellular homology directly (but, again, without the cellular boundary formula). I don't know if my notation (which is from Rotman's book) is standard, so I'll mention it here: For a CW complex $X$, I write $X^k_Y=X^{(k)}\cup Y$, where $X^{(k)}$ is the $k$-skeleton. I write $W_*(X,Y)$ for the chain complex where $W_k(X,Y)=H_k(X^k_Y,X^{k-1}_Y)$.

I know that $H_k(W_*(X,Y))\cong H_k(X,Y)$. Set $X=T$ and $Y=\emptyset$. Then I have shown that the chain $W_*(T,\emptyset)$ looks like $$W_3(T)=0\to\mathbb Z\to\mathbb Z\oplus\mathbb Z\to\mathbb Z\to0=W_{-1}(T).$$ Unfortunately, I don't really know what the differentiation maps $d_n:W_n(T)\to W_{n-1}(T)$ look like. Also, this method doesn't seem to use the part of the hint about $H_1(T,S^1\vee S^1)$, so my guess is that this isn't the intended solution.

Any suggestions for how to find the homology groups would be appreciated! (Also, if my notation is weird, feel free to change it.)

Answer 1

You can compute it without using the cellular boundary formula by using simplicial homology instead of cellular homology. The torus has the following $\Delta$-complex structure:

Here $x$ is a $0$-cell, $a$,$b$,$c$ are $1$-cells, and $\rho_1$,$\rho_2$ are $2$-cells. We compute the differentials: \begin{equation*} \begin{split} d_1(a)&=x-x=0 \\ d_1(b)&=x-x=0 \\ d_1(c)&=x-x=0 \end{split} \hspace{3cm} \begin{split} d_0(x)&=0 \\ d_2(\rho_1)&=a+b-c \\ d_2(\rho_2)&=a+b-c. \end{split} \end{equation*} We then see that the simplicial chain complex becomes: $$ \cdots \to 0 \longrightarrow \underset{(2)}{\mathbb{Z}\{\rho_1,\rho_2\}} \xrightarrow{ \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ -1 & -1 \end{bmatrix}} \underset{(1)}{\mathbb{Z}\{a,b,c\}} \overset{0}{\longrightarrow} \underset{(0)}{\mathbb{Z}\{x\}} \longrightarrow 0 \to \cdots. $$ Taking homology of this chain complex yields the homology groups of the torus as wanted.