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When computing relative homology, I know how to find the maps, how to find all the absolute homology groups, and how to show $H_2(X,A)$ is (usually) $0$. My issue is I do not understand how to compute $H_1$ and $H_0$ in this example, but for different reasons than the OP. I do not see where this comes from:

"I use the theorem that says that the succession of $$0\to H_1(\partial M)\to H_1(M)\to H_1(M,\partial M)\to 0$$ is an exact one."

Well, if that is true, $H_1(M,\partial M)$ is immediate, because for any exact $0 \to A \stackrel{f}\to B \to C \to 0$ we have $C\cong B/\text{im}(f)$, and I already know the image there is $2\mathbb{Z}$. I can find $H_0$ similarly.

But I am unfamiliar with this idea of isolating a smaller short exact sequence from a long exact sequence. When can one do this and why?

In the LES for $H_n(M,\partial M)$, the group $H_1(M,\partial M)$ is followed by $\mathbb{Z} \to \mathbb{Z} \to H_0(M,\partial M) \to 0$. Why can we "skip" all this information and "jump" to the last $0$ to obtain the SES above? Or is that not what's going on here?

contingent
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  • Are $M$ and $\partial M$ connected? – Paul Frost Aug 14 '22 at 17:43
  • Yes, but I don't know what this tells me about relative homology. – contingent Aug 14 '22 at 17:45
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    The answer to the question that you cite explains things: since $M$ and $\partial M$ are path connected, the induced map on $H_0$ is an isomorphism, so the map $H_1(M, \partial M) \to H_0(\partial M)$ is the zero map. The answer also explains why the map $H_1(\partial M) \to H_1(M)$ is injective, and this implies that the preceding map in the long exact sequence is zero. – John Palmieri Aug 14 '22 at 18:46

1 Answers1

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The easiest way to see it is to use the long exact sequence of reduced homology groups. See for example Definition of $\tilde{H}_n(X,A)$.

Since $\partial M$ is path connected, we have $\tilde H_0(\partial M) = 0$.

If you do not want to use reduced homology groups, you can argue as follows.

$i_* : H_0(\partial M) \to H_0(M)$ is an injection (note 1. $H_0(X)$ is generated by the path components of $X$ 2. The single path component of $\partial M$ is mapped to a unique path component of $M$. This argument does not require $M$ path connected, it may have arbitrarily many path components).

Now look at $H_1(M) \stackrel{j_*}{\to} H_1(M,\partial M) \stackrel{\partial}{\to} H_0(\partial M) \stackrel{i_*}{\to} H_0(M)$. We have $\operatorname{im} \partial = \ker i_* = 0$, thus $\partial$ is the zero map. We conclude that $\operatorname{im} j_* = \ker \partial = H_1(M,\partial M) $, i.e. $j_*$ is a surjection.

Paul Frost
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