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I am not sure if my approach is on the right track for $\Rightarrow$, since I don't know how to proceed and reach the conclusion that $A$ meets each path-component of $X$? And I hope my approach to the other direction make sense?

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

By Theorem 2.16, we have the exact sequence:

$$\cdots H_0(A) \stackrel{i_*}{\to} H_0(X) \stackrel{j_*}{\to} H_0(X,A) \to 0$$

$\Rightarrow$: If $H_0(X, A) = 0$, then $H_0(X) = \ker j_*(H_0(X)) = \operatorname{Im}i_*(H_0(A))$. Because $H_0(X) = \operatorname{Im}i_*(H_0(A))$, we know $i_*$ is surjective.

Edit: I hope I figured this out?

The map $i_*: H_0(A) \to H_0(X)$ is surjective implies that the map $i_*: \bigoplus_\alpha H_0(A_\alpha) \to \bigoplus_\beta H_0(X_\beta)$ is surjective. That is to say that $\forall \beta, \exists A_\alpha : i(A_\alpha) \subset X_\beta$. Hence $A$ meets each path-component of $X$.

$\Leftarrow$: According to Proposition 2.7, for any space $X$, $H_0(X)$ is a direct sum of $\mathbb{Z}$'s, one for each path-component of $X$. So if $A$ meets each path-component of $X$, $H_0(X) = H_0(A) \approx \mathbb{Z}$. So the inclusion $i: A \to X$ induces a surjection $H_0(A) \to H_0(X)$.

So since $H_0(A) \to H_0(X)$ is surjective, $\operatorname{Im} i_*(H_0(X)) = H_0(X, A) = \ker H_0(X, A)$. Since the entire $H_0(X, A)$ lies in the kernel of $j_*$, $H_0(X, A) = 0$.

Stefan Hamcke
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1LiterTears
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  • Oops, I mean Hatcher. Sorry @nik. – 1LiterTears Nov 03 '13 at 22:12
  • You should be careful with your notation. You take the image and kernel of homomorphisms, not groups. So for instance your last line should read '$\ker j_=\mbox{Im }i_=H_0(X)$. Since the whole group $H_0(X)$ lies in the kernal of $j_$, we must have $0=\mbox{Im }j=\ker(H_0(X,A)\rightarrow 0)$, and so $H_0(X,A)=0$ as $H_0(X,A)\rightarrow 0$ is surjective and so is also isomorphic.' – Dan Rust Nov 03 '13 at 22:44
  • Oh that's so right. Thank you very much @DanielRust! – 1LiterTears Nov 05 '13 at 15:02

1 Answers1

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For the left to right, you definitely started off correctly by showing that $i_*$ is surjective. Now the next step to show is that this implies for each path component $P_j(X)$, we have $P_j(X)\cap A\neq \emptyset$. A set of generators for $H_0(X)$ is given by the homology classes represented by the $0$-simplices $p_j\colon\{*\}\rightarrow X$ given by $p_j(*)=x_j$ with $x_j\in P_j(X)$. Note that the homology class $[p_j]\in H_0(X)$ is given precisely by the $0$-simplices whose image is in $P_j(X)$ - that is, two simplices are homologous if and only if their images lie in the same path component.

Let $f\colon\{*\}\rightarrow A$ by a representative of a generator of $H_0(A)$ such that $i_*[f]=[p_j]$ for some $j$, which is guarenteed to exist by the surjectivity of $i_*$. This tells use that $[i\circ f]=[p_j]$ and so $i\circ f$ is homologous to $p_j$. It follows that the image of $i\circ f$ lies in the same path component as the image of $p_j$ and so $i(f(*))\in P_j$. But $i$ is just inclusion and so $f(*)\in P_j$. Given that $f(*)$ is an element of $A$, this tells us that there is some element in both $A$ and $P_j(X)$ hence $P_j(X)\cap A\neq\emptyset$ as required.

For the second part, you're nearly right, except you said each $0$th homology group is isomorphic to $\mathbb{Z}$ which isn't always the case for non-path-connected space (we don't even need $H_0(A)\cong H_0(X)$ in fact - consider $A=\{0,1\}\subset X=\mathbb{R}$). The important part here is to show that if $A$ meets each path component in $X$, then we can pick an element $y_j\in A$ in each path component $P_j(X)$ and then the image of the $H_0(A)$ homology classes represented by the simplices which have the $y_i$ as their image, under $i_*$, is the whole of $H_0(X)$ because $H_0(X)$ is generated by simplices whose image lies in the path components $P_j(X)$. This then implies surjectivity and the rest of your proof follows.

Dan Rust
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