Homology Groups

**Defunkt** · December 6th 2010, 07:09 AM

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 $X/A$ is the wedge of two tori, that $H_i(X,A)=\tilde{H}_i(X/A) = \left\{ \begin{array}{lr} \mathbb{Z}^2 & i=0\\ \mathbb{Z}^4 & i=1\\ \mathbb{Z}^2 & i=2 \end{array} \right.$

**Drexel28** · December 13th 2010, 12:36 PM

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 $X/A$ is the wedge of two tori, that $H_i(X,A)=\tilde{H}_i(X/A) = \left\{ \begin{array}{lr} \mathbb{Z}^2 & i=0\\ \mathbb{Z}^4 & i=1\\ \mathbb{Z}^2 & i=2 \end{array} \right.$

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 $A$ in $H_n(X,A)$ ? 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 $\mathbb{T}^2$ into a zero-cell, two one-cells, and one two-cell in the natural way so that our chain complex becomes $0\to\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}\to\mat hbb{Z}\to 0$ . I think (giving the outward normal orientation of course) it's easy to show that $\partial_0:\mathbb{Z}\to 0,\partial_1:\mathbb{Z}\oplus\mathbb{Z},\partial_2 :\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$ and $\partial_3:0\to\mathbb{Z}$ are all trivial. From there it's pretty easy to calculate that

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

**Defunkt** · December 13th 2010, 01:33 PM

Haven't gone through cellular homology yet, so only used singular homology, exact sequences and excision for these:

for a), note that $(X,A)$ is a good pair, and so $H_n(X,A) \cong \tilde{H}_n(X /A)$ . The quotient's 1-skeleton has only one path-component, and therefore $H_0(X,A) \cong \tilde{H}_0(X/A) = 0$ .

For n=1, use the long exact sequence of the pair X,A:

$...\rightarrow \tilde{H}_1(A) \rightarrow \tilde{H}_1(S^2) \rightarrow H_1(S^2,A) \rightarrow \tilde{H}_0(A) \rightarrow \tilde{H}_0(S^2)\rightarrow ...$

$\tilde{H}_1(S^2) = 0$ and $\tilde{H}_0(S^2) = 0$ and $\tilde{H}_0(A) = \mathbb{Z}^{|A|-1}$ . The first two imply that the map $H_1(X,A) \rightarrow \tilde{H}_0(A)$ is an isomorphism and from the latter we have the result for n=1.

For n=2, $S_2(A) = 0$ and therefore $H_2(S^2,A) = H_2(S^2) = \mathbb{Z}$ .

For any n>2, the result is obviously 0.

Similar arguments give us the same $H_0(X,A),H_2(X,A)$ for $X=S^1 \times S^1$ . For the first homology group, the long exact sequence we used to calculate $H_1(S^2,A)$ is
$...\rightarrow 0 \rightarrow \mathbb{Z}^2 \rightarrow H_1(X,A) \rightarrow \mathbb{Z}^{|A|-1} \rightarrow 0 \rightarrow ...$
and so we get $H_1(S^1 \times S^1, A) = \mathbb{Z}^{|A|+1}$

To finish (b), we note that $X/B$ is homotopy equivalent to a torus with 2 points identified, and since we already know that homotopic spaces have the same homology groups, and since $(X,B)$ is a good pair, we get:

$H_n(X,B) \cong \tilde{H}_n(X/B) \cong \tilde{H}_n(\mathbb{T}^2/{\{x_0,y_0\}) \cong {H}_n(\mathbb{T}^2, \{x_0,y_0\})$

But we already calculated this one in (a).

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