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.)

  • $\begingroup$ There are many ways to approach this, but one would be to consider that we have a map $T^2\to S^2$ given by composing the characteristic map of the $2$-cell and collapsing the $1$-skeleton to a point. This is clearly equal to the characteristic map of the $2$-cell of $S^2$ (built with one $0$-cell and one $2$-cell), so takes the generator of $H_2T^2$ to $H_2S^2$. $\endgroup$
    – pancini
    Apr 13, 2021 at 21:59
  • $\begingroup$ @ElliotG Thanks! Maybe I should've mentioned that this is something like the third or fourth time that Rotman's asked us to compute the homology groups of $T$, so I've actually already done it that way :P $\endgroup$
    – boink
    Apr 13, 2021 at 22:05
  • $\begingroup$ Do $S^1$ and $S_1$ both denote 1-spheres? $\endgroup$
    – memerson
    Apr 13, 2021 at 22:28
  • $\begingroup$ @memerson whoops, that’s a typo! They should both be the 1-sphere $S^1$ $\endgroup$
    – boink
    Apr 13, 2021 at 22:29

1 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.

  • $\begingroup$ Just FYI: I didn't accepted the solution just because I was looking for a solution with cellular homology—I've already found the homology groups with simplicial homology. Thanks for the help though! $\endgroup$
    – boink
    Apr 18, 2021 at 3:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .