Originally Posted by

**Defunkt** Hey, I've been reading Hatcher and doing the exercises, couldn't answer this one:

(p.132 ex. 17)

Tried various directions but I'm pretty much lost. Any help will be appreciated!

Edit: I did get that since $\displaystyle X/A$ is the wedge of two tori, that $\displaystyle H_i(X,A)=\tilde{H}_i(X/A) = \left\{

\begin{array}{lr}

\mathbb{Z}^2 & i=0\\

\mathbb{Z}^4 & i=1\\

\mathbb{Z}^2 & i=2

\end{array}

\right.$

I am very, very novice at homology theory so take what I am about to say as probably wrong (and possibly gobbeldy-gook). That said, since no one has answered your question I'll take a stab at my interpretation (see below) of it.

So, what is the $\displaystyle A$ in $\displaystyle H_n(X,A)$? Is it the cellular decomposition (I have Hatcher, but not with me. And, I haven't read it yet)? If so, have you discussed the fact yet that the homology is independent of the decomposition? If you have then I think I can help with a) (maybe). Why can't you decompose $\displaystyle \mathbb{T}^2$ into a zero-cell, two one-cells, and one two-cell in the natural way so that our chain complex becomes $\displaystyle 0\to\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}\to\mat hbb{Z}\to 0$. I think (giving the outward normal orientation of course) it's easy to show that $\displaystyle \partial_0:\mathbb{Z}\to 0,\partial_1:\mathbb{Z}\oplus\mathbb{Z},\partial_2 :\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$ and $\displaystyle \partial_3:0\to\mathbb{Z}$ are all trivial. From there it's pretty easy to calculate that

$\displaystyle H_n(\mathbb{T}^2)=\begin{cases}\mathbb{Z} & \mbox{if}\quad n=0,2\\ \mathbb{Z}\oplus\mathbb{Z} & \mbox{if}\quad n=1\\ 0 & \mbox{otherwise}\end{cases}$