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I've been thinking about the exactness "axiom" which says that

For any topological pair $(X,A)$ the following sequence is exact:

$$ \dots \rightarrow H_n(X) \rightarrow H_n(X,A) \rightarrow H_{n-1}(A) \rightarrow \dots$$

I thought I could use this to compute the relative homology so I tried the following example:

$H_n(D^2, S^1) = ?$

By exactness the following sequence is exact:

$$ \dots \rightarrow H_n(D^2) \xrightarrow{j_\ast} H_n(D^2,S^1) \xrightarrow{\partial_\ast} H_{n-1}(S^1) \rightarrow \dots$$

Then I though I could use the following 2 facts to find $H_n(D^2, S^1) $:

$H_n(D^2) = H_n(\{\ast\}) = 0 (n>0) $ and $\mathbb{Z}(n=0)$

$H_n(S^1) = 0 (n \geq 2) $ and $\mathbb{Z}(n=0,1)$

But somehow knowing $im j_\ast = ker\partial_\ast$ etc. doesn't give me any useful information to find $H_n(D^2, S^1) $. What am I doing wrong? Is this not a good example of when to apply the exactness axiom? If no, can someone please show me a better one? Many thanks for your help!


OK, after Theo's comment I produced the following exact sequence:

$$ \dots 0 \xrightarrow{a} H_2(D^2, S^1) \xrightarrow{b} \mathbb{Z} \rightarrow 0 \xrightarrow{c} H_1(D^2, S^1) \xrightarrow{d} \mathbb{Z} \rightarrow \mathbb{Z} \xrightarrow{e} H_0(D^2, S^1) \xrightarrow{f} 0$$

I assume that $H_n(D^2, S^1) = 0$ for $n > 2$ because the space doesn't have any $n$ cells in it but I'm not entirely sure that's rigorous. Anyway, with the exactness of the sequence I get the following:

(i) $im a = 0 = ker b $ $\implies b$ is injective

(ii) $ker f = H_0(D^2,S^1) = im e$

(iii) $im c = 0 = ker d$ $\implies d$ injective

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You should at least be able to compute $H_2(D^2,S^1)$. You can't have much more useful information than knowing that $\partial_\ast$ is an isomorphism! –  t.b. Aug 15 '11 at 10:04
You don't need to know anything about $n$ cells. For $n > 2$, the fact that $H_n(D^2,S^1) = 0$ is a consequence of that group being between two zeros in an exact sequence. –  a.r. Aug 15 '11 at 10:29
Are you sure about the domain of $c$? –  t.b. Aug 15 '11 at 10:31
Oh noes, that's a typo. I have it right further above, thanks for pointing it out! –  Matt N. Aug 15 '11 at 10:46
Essentially every textbook shows how to get Mayer-Vietoris from that exact sequence, and then you can compute. –  Mariano Suárez-Alvarez Aug 16 '11 at 6:01
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3 Answers

up vote 6 down vote accepted

Since $H_n(D^2) = H_n(S^1) = 0$ for $n \geq 2$, we also have $H_n(D^2, S^1) = 0$ for $n> 2$. So we must focus on $H_n(D^2,S^1)$ for $n=0,1,2$.

For $n=0$, we have

$$ \dots \longrightarrow H_0(S^1) \longrightarrow H_0(D^2) \longrightarrow H_0(D^2,S^1) \longrightarrow 0 $$

which is

$$ \dots \longrightarrow \mathbb{Z} = \mathbb{Z} \longrightarrow H_0(D^2,S^1) \longrightarrow 0 \ . $$


$$ H_0(D^2,S^1) = 0 \ . $$

For $n=1$, the situation is

$$ \dots \longrightarrow H_1(D^2) \longrightarrow H_1(D^2,S^1) \longrightarrow H_0(S^1) \longrightarrow H_0(D^2) \longrightarrow \dots $$

That is,

$$ 0 \longrightarrow H_1(D^2,S^1) \longrightarrow \mathbb{Z} = \mathbb{Z} $$

Which again means

$$ H_1(D^2, S^1) = 0 \ . $$

Finally, for $n=2$,

$$ \dots \longrightarrow H_2(D^2) \longrightarrow H_2(D^2,S^1) \longrightarrow H_1(S^1) \longrightarrow H_1(D^2) \longrightarrow \dots $$

Which is

$$ 0 \longrightarrow H_2(D^2, S^1) \longrightarrow \mathbb{Z} \longrightarrow 0 $$


$$ H_2(D^2, S^1) = \mathbb{Z} \ . $$

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Where $H_0(S^1) \to H_0(D^2)$ being an isomorphism requires some geometric insight on the inclusion $S^1 \to D^2$. –  Alexander Thumm Aug 15 '11 at 10:33
@Alexander. Yes. I took it for granted. If you have a path-connected space such as $S^1$, then any point $x\in S^1$ can be taken as a generator of $H_0(S^1)$. The same point can be taken as a generator of $H_0(D^2)$. So the inclusion $S^1 \rightarrow D^2$, with these generators, induces the identity on homology. –  a.r. Aug 15 '11 at 10:38
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Well, you know that that sequence up there is exact. Okay, what does that mean? Well, so we have a long trail of 0s, and remember that a sequence of the form: $0 \rightarrow A \rightarrow^f B$ is exact if and only if f is injective. So, we have that in this case you'll have a long trail of 0s, followed by: $0 \rightarrow H_2(D^2,S^1) \rightarrow^{\partial} H_1(S^1) \rightarrow 0 \cdots$. Well, so we have that $\partial$ is injective, and surjective, that is, an isomorphism. So $H_2(D^2,S_1) = Z$. And in the same way, we can reason for the other terms. Note that we used that a sequence $0 \rightarrow A \rightarrow^f B \rightarrow 0$ is exact iff f is surjective and injective.

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Using your helpful answers and comments I'm posting the answer in my own words, including the details I was missing (knew them but failed to apply):

Fact 1: A group homomorphism $\varphi: G \rightarrow G^\prime$ where $G = <g>$ and $G^\prime = <g^\prime>$ such that $\varphi(g) = g^\prime$ is an isomorphism.

Fact 2: If $X$ is path-connected any two constant singular 0-simplexes $\sigma, \sigma^\prime$ in $C_0(X)$ differ by a boundary, i.e. $\exists$ a path $ \gamma \in C_1(X)$ such that $\partial \gamma = \sigma + \sigma^\prime$. Therefore they are all in the same homology class and therefore any constant singular 0-simplex in $X$ generates $H_0(X)$.

Fact 3: As a consequence of facts 1 and 2, the inclusion $i: A \hookrightarrow X$ induces an isomorphism $i_\ast: H_0(A) \rightarrow H_0(X)$.

Case $n = 0$:

$$ \dots \rightarrow H_0(S^1) \xrightarrow{i_\ast} H_0(D^2) \rightarrow H_0(D^2, S^1) \rightarrow 0$$


$$ \dots \rightarrow \mathbb{Z} \xrightarrow{i_\ast} \mathbb{Z} \xrightarrow{f} H_0(D^2, S^1) \xrightarrow{g} 0$$

Now because $i_\ast$ is an isomorphism, $im i_\ast = \mathbb{Z} = ker f$ $ \implies f = 0$ and therefore $im f = 0 = ker g$, so $g$ is injective. But $g = 0$ is also surjective, so $g$ is an isomorphism and therefore $H_0(D^2, S^1) \cong 0$.

Case $n=1$

$$ \rightarrow H_1(S^1) \xrightarrow{i_\ast}  H_1(D^2) \xrightarrow{f} H_1(D^2, S^1) \xrightarrow{\partial_\ast} H_0(S^1) \xrightarrow{i_\ast} H_0(D^2) \dots$$

where $H_1(S^1) = \mathbb{Z}$, $H_1(D^2) = 0$, $H_0(S^1) = \mathbb{Z} = H_0(D^2)$

(i) $im i_\ast = 0 = ker f $ $\implies f$ injective

(ii) $ker i_\ast = 0 = im \partial_\ast $ $\implies \partial_\ast = 0$ (const.)

(iii) $ker \partial_\ast = H_1(D^2, S^1) = im f $ $\implies f$ surjective

$\implies f$ is isomorphism

$\implies H_1(D^2, S^1) = 0$

Case $n=2$

$$ 0 \xrightarrow{f} H_2(D^2, S^1) \xrightarrow{g} \mathbb{Z} \xrightarrow{h} 0$$

(i) $im f = 0 = ker g $ $\implies g$ injective

(ii) $im h = 0 \implies ker h = \mathbb{Z} = im g \implies g$ surjective

$\implies g$ is isomorphism

$\implies H_2(D^2, S^1) \cong \mathbb{Z}$

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I think I don't understand how to get an isomorphism or the identity on the homology groups. How does the inclusion $S^1 \rightarrow D^2$ induce the identity on the homology groups? –  Matt N. Aug 16 '11 at 5:39
Do you agree with me that any point $x\in S^1$ generates $H_0(S^1) \cong \mathbb{Z}$? Do you agree with me that any point $x \in D^2$ generates $H_0(D^2) \cong \mathbb{Z}$? Then, what is the image of $x\in S^1$ by the inclusion $S^1 \rightarrow D^2$? –  a.r. Aug 16 '11 at 6:40
I insist that any point $x\in X$ in a path-connected space $X$ generates $H_0(X)\cong \mathbb{Z}$, but NOT $C_0(X)$ which is isomorphic to $\bigoplus_{x\in X} \mathbb{Z}\langle x \rangle$, where $\mathbb{Z}\langle x \rangle\cong \mathbb{Z}$ is the free (abelian) group generated by the point $x$. –  a.r. Aug 16 '11 at 6:43
Yes, many thanks for correcting me. Now I understand better. I corrected my answer. –  Matt N. Aug 16 '11 at 6:58
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