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I've been doing some more exercises in Hatcher, in particular the following:

Show that $H_0(X,A) = 0$ iff $A$ meets each path-component of $X$.

"$\Leftarrow$": Let $x_i \in A \cap X_i \neq \emptyset \forall $ path-components $X_i$. Then for any $x \in X_i$ there is a path $\gamma$ from $x_i$ to $x$. They therefore differ by a boundary: $\partial \gamma = x - x_i$ and therefore $\{ x \} = \{ x_i \}$ in $H_0(X,A) $. Therefore $H_0(X,A) = 0$.

"$\Rightarrow$" Now for this direction I'm not so sure and I'd be glad if you could give me a hint. Let $H_0(X,A) = 0$. By definition, the relative homology group is calculated from the following sequence of chain groups:

$$ 0 \rightarrow C_0(A) \rightarrow C_0(X) \xrightarrow{\partial_1} C_0(X) / C_0(A) \xrightarrow{\partial_0} 0$$

Then $H_0(X, A) = H_0(C_0(X) / C_0(A)) = ker \partial_0 / im \partial_0 = (C_0(X) / C_0(A) ) / im \partial_1 = 0$.

There are two cases where this equality holds:

(i) $C_0(X) / C_0(A) = 0$

(ii) $im \partial_1 = C_0(X) / C_0(A)$

In case (i), $A \cap X_i \neq \emptyset$ for all $X_i$.

In case (ii) I'm not sure how to proceed. Can you give me a hint? Many thanks for your help!

Edit Or maybe I could do the second direction like this:

$$0 = H_0(X,A) = \oplus_i H_0(X_i, A) \implies H_0(X_i, A) = 0 \forall i$$

$$ \implies A \cap X_i \neq \emptyset \forall i$$?

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The relative homology group is not computed from the sequence you mention... – Mariano Suárez-Alvarez Sep 5 '11 at 7:07
But they mention it on Wikipedia: They don't explain why so I assumed they use it to compute $H_n(X,A)$. Why do they mention it there? – Rudy the Reindeer Sep 5 '11 at 7:20
Matt, you should probably read the exposition in Hatcher about relative homology, rather than Wikipedia. It will be immensely more helpful. – Mariano Suárez-Alvarez Sep 5 '11 at 7:21
Yeah, I just looked at it. But then why do they mention said sequence on Wikipedia in that context? – Rudy the Reindeer Sep 5 '11 at 7:27
Matt: nowhere in that page do they say that relative homology is computed as you say it is. They mention that sequence because it defines the complex $C_\bullet(X)/C_\bullet(A)$ whose homology is $H_\bullet(X,A)$. – Mariano Suárez-Alvarez Sep 5 '11 at 7:29
up vote 7 down vote accepted

You can reduce the problem before you start, as follows: Suppose the path components of $X$ are the spaces $X_i$ with $i\in I$. For each $i$, let $A_i=X_i\cap A$, and show that $$H_p(X,A)\cong\bigoplus_{i\in I}H_p(X_i,A_i).$$ Using this, you are left with proving the statement

If $X$ is a non-empty path connected space and $A\subseteq X$, then $H_0(X,A)=0$ iff $A$ is not-empty.

One way to prove this is to consider the end of the long exact sequence for the pair $(X,A)$, namely $$H_0(A)\to H_0(X)\to H_0(X,A)\to 0$$ By our hypothesis, $H_0(X)\cong\mathbb Z$, and you should know/check that it is generated by the homology class of any point in $X$. If $A$ is empty, then exactness immediately tells you wat $H_0(X,A)$ is non-zero. If $A$ is non-empty, pick a point $a\in A$ and consider the homology class $[a]\in H_0(A)$. The image of $[a]$ under $H_0(A)\to H_0(X)$ is the homology class of a point, which generates the codomain. Exactness now implies that $H_0(X,A)=0$.

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Thank you! I think I might have found something shorter: I could use the previous question in Hatcher which states that $H_n(X,A) = 0 \iff i:A\hookrightarrow X$ induces an isomorphism.... – Rudy the Reindeer Sep 5 '11 at 7:21
If $i_\ast$ is an isomorphism $A$ has to meet all path components of $X$. – Rudy the Reindeer Sep 5 '11 at 7:22
@Matt: that does not work, because we do not know that the inclusion induces an isomorphism in all degrees---and, in fact, it may well not do that! – Mariano Suárez-Alvarez Sep 5 '11 at 7:23
But that's what I proved in question 15 on page 114: if $H_n(X,A) = 0$ for all $n$ then $i_\ast$ is an isomorphism for all $n$! – Rudy the Reindeer Sep 5 '11 at 7:27
@Matt: in this context, we have no information about what is happening for $n>0$. Consider the inclusion of a sphere $S^2$ into $\mathbb R^3$: it does not induce isomorphisms for all $n$. – Mariano Suárez-Alvarez Sep 5 '11 at 7:28

I'm not sure if this answer is correct but I'll give it a try. Suppose that $H_0(X,A) = 0$. Then we know that $i_\ast H_0(A) \hookrightarrow H_0(X)$ is surjective. Now choose some $x \in X$ that is in some path component $x_i$. Then we can consider an associated singular $0$ - simplex $\sigma : \Delta^0 \to X$ that maps $\Delta^0$ to $x$. Surjectivity of $i_\ast$ implies that I can find a $\tau : \Delta^0 \to A$ such that

$$[\tau \circ i] = [\sigma]$$ so that $\tau \circ i- \sigma$ is the boundary of a singular $1$ - chain in $X$. Now I think from here you can say that this singular $1$ - chain is actually a singular $1$ - simplex $\rho$. Since the image of $\rho$ is always path - connected, we have that the image of $i \circ \tau$ and $\sigma$ are both in the same path - component. Since $\sigma : \Delta^0 \to x$ with $x$ in an arbitrary path - component, we are done.

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hello, please why $i_*$ is surjective ? – Vrouvrou Nov 9 '13 at 21:51

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