# Relative homology groups of the torus

I have the following question to problem 2.1.17 in Allen Hatcher's "Algebraic Topology". So far I came up with the following exact sequences (for A and B):

\begin{aligned} 0&\rightarrow H_{2}(A) \rightarrow H_{2}(X) \rightarrow H_{2}(X,A)\rightarrow\\ &\rightarrow H_{1}(A) \rightarrow H_{1}(X) \rightarrow H_{1}(X,A)\rightarrow\\ &\rightarrow H_{0}(A) \rightarrow H_{0}(X) \rightarrow H_{0}(X,A) \rightarrow 0 \end{aligned} and \begin{aligned} 0&\rightarrow H_{2}(B) \rightarrow H_{2}(X) \rightarrow H_{2}(X,B)\rightarrow\\ &\rightarrow H_{1}(B) \rightarrow H_{1}(X) \rightarrow H_{1}(X,B)\rightarrow\\ &\rightarrow H_{0}(B) \rightarrow H_{0}(X) \rightarrow H_{0}(X,B) \rightarrow 0, \end{aligned} where $H_{2}(A) = H_{2}(B) = 0$, $H_{1}(A) = H_{1}(B) = \mathbb{Z} = H_{0}(A) = H_{0}(B)$ and for $X$ there is $H_{2}(X) = H_{0}(X) = \mathbb{Z}$ and $H_{1}(X) = \mathbb{Z}^{4}$. Furthermore I know that the mappings $H_{1}(A) \rightarrow H_{1}(X)$ is zero and that $H_{1}(B) \rightarrow H_{1}(X)$ is injective. By these I could deduce that $H_{0}(X,A) = 0$ and $H_{1}(X,A) = \mathbb{Z}^{4}$ and $H_{0}(X,B) = 0$. But I can't go on further. What about the other relative homology groups? What do I need more? Hope this question is not too trivial and apologize. Hope someone to help.

mika

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I would be greatly obliged if you were to include 2.1.17 from AT in this question. But I greatly appreciate you showing your work so far. –  mixedmath Dec 31 '11 at 8:04
my questions are: are my calculations so far correct? How can I get more information on the other relative homology groups? –  beno Dec 31 '11 at 8:07
@mika I think mixedmath was asking you to say what you're trying to find and what $X, A, B$ are and so on. Not everyone has the book in front of them, and while Hatcher's book is freely available I think it's best for questions to be roughly self-contained. Looking at the book it seems like you're doing part (b) of the problem; is this correct? –  Dylan Moreland Dec 31 '11 at 8:11
yes, I do part (b). –  beno Dec 31 '11 at 8:15

## 2 Answers

$(X,A)$ and $(X,B)$ are good pairs (because $X$ is a cell complex and $A$ and $B$ are subcomplexes).

Now you can use proposition 2.22. on page 124 which states that $H_n(X,A) \cong \tilde{H_n}(X/A)$.

In your case you have $X/A = T^2 \vee T^2$ and $X/B = T^2 \vee S^1$. So you want to compute the reduced homology of a wedge sum. By corollary 2.25. on page 126 you know that $\tilde{H_n} (\bigvee_\alpha X_\alpha) = \bigoplus_\alpha \tilde{H_n}(X_\alpha)$ so the answer to the question boils down to computing the reduced homology groups of $T^2$ and $S^1$ respectively.

Hope this helps. Otherwise don't hesitate to ask.

As for your question: It would be nice of you if you could edit it and include the question from Hatcher. Would you do that? Thank you in advance.

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And also clean up the numerous spelling errors and formatting issues. –  user641 Dec 31 '11 at 9:24

For the case $H(X,A)$: just plug in the exact sequence the things you know:

$0 \to \mathbb{Z} \to H_2(X,A) \to \mathbb{Z} =H_1(A)\to H_1(X)= \mathbb{Z}^4 \to H_1(X,A)\to 0$

As you said, the map $\mathbb{Z} =H_1(A)\to H_1(X)= \mathbb{Z}^4$ is the zero map since $A$ bounds a subsurface in $X$; thus you can split your sequence in two easier pieces:

$0 \to \mathbb{Z} \to H_2(X,A) \to \mathbb{Z} =H_1(A)\to 0$

$0 \to H_1(X)= \mathbb{Z}^4\to H_1(X,A)\to 0$

But these are exact sequences! so you obtain respectively

$H_2(X,A) / \mathbb{Z} = H_1(A)$, which implies $H_2(X,A)=\mathbb{Z} ^2$

$H_1(X,A)=\mathbb{Z}^4$

Now take the case (X,B) and plug in what you know

$0 \to \mathbb{Z} \to H_2(X,B) \to \mathbb{Z} =H_1(B)\to H_1(X)= \mathbb{Z}^4 \to H_1(X,B)\to 0$

Now $B$ is not nullhomologous and so the map $H_1(B) \to H_1(X)$ is injective; therefore, you can split like this:

$0 \to \mathbb{Z} \to H_2(X,B) \to 0$ which implies $H_2(X,B)=\mathbb{Z}$

$0\to \mathbb{Z} \to \mathbb{Z}^4 \to H_1(X,b)$ which implies $H_1(X,B) = \mathbb{Z}^4 / \mathbb{Z} = \mathbb{Z}^3$

Notice that:

1) this is consistent with Matt's answer, but is more basic: you don't need to find any retract of $X/A$ or $X/B$ and you don't need the result on wedge sums, just some reasoning on injective maps between powers of $\mathbb{Z}$.

2) this generalizes to a surface of arbitrary genus $g$ (you will have $2g$ instead of $4$, but the proof is exactly the same).

Let me know if something was not clear.

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