I am trying to do the stated problem in Hatcher:

Show $H_1(X,A) = 0$ iff $H_1(A) \to H_1(X)$ is surjective and each path component of $X$ contains at most one path component of $A$.

Now I have reduced the problem to showing that $i_\ast : H_0(A) \to H_0(X)$ injective iff each path component of $X$ contains at most one path component of $A$. This comes from looking at the end of the LES of the pair $(X,A)$:

$$\ldots \to H_1(X) \to H_1(X,A) \to H_0(A) \to H_0(X) \to H_0(X,A) \to 0$$

Now one direction I have shown, the other that is giving me trouble is the converse. That is if $i_\ast$ is not injective then there is a path component of $X$ that contains at least two path components of $A$. I have the following:

Suppose $i_\ast$ is not injective. Then there is a $\tau \in C_0(A)$ such that $[\tau \circ i] = 0$ but $[\tau] \neq 0$. That is to say, $\tau \circ i = \partial(\sigma)$ for some $\sigma \in C_1(X)$ but $\tau$ is not the boundary of any $\sigma'\in C_1(A)$. However I'm confused because to me the only way for $\tau \circ i$ to be the boundary of a singular $1$ - simplex $\sigma$ in $X$ is if $\sigma$ is a loop. What's wrong here?


  • $\begingroup$ Why do you assume $\sigma'\in H_1(A)$ in contrast to $\sigma\in C_1(X)$? Homology classes are always represented by cycles, so this would mean that $\partial(\sigma')=0$. If it's just a typo, maybe you could correct it along with the map in the quote which is $H_1(A)\to H_1(X)$. $\endgroup$ – Stefan Hamcke Nov 10 '12 at 17:41
  • $\begingroup$ @StefanH. I have corrected the first point in the comment. I don't understand which map you're talking about in the second. $\endgroup$ – user38268 Nov 10 '12 at 22:40
  • $\begingroup$ Why must $\sigma$ be a loop? Are you assuming that $\tau$ is a single vertex rather than an arbitrary 0-chain in $A$? $\endgroup$ – John Palmieri Nov 10 '12 at 23:49
  • $\begingroup$ @BenjaLim. I was talking about the map $H_1(A)\to H_1(X)$ in the quotation of the problem, where you switched $A$ and $X$. $\endgroup$ – Stefan Hamcke Nov 11 '12 at 13:06
  • $\begingroup$ @StefanH. I have corrected it now. Thanks. $\endgroup$ – user38268 Nov 11 '12 at 23:40

So first, as you have remarked, if $H_1(X,A) = 0$ then it is immediate from the long exact sequence that $H_1(A) \to H_1(X)$ is surjective and that $H_0(A) \to H_0(X)$ is injective. Conversely, suppose that $H_1(X,A) \neq 0$, so let $\sigma \neq 0 \in H_1(X,A)$. Then you have two possibilities:

  • Either $\partial \sigma = 0 \in H_0(A)$. Then by exactness, there exists $\gamma \in H_1(X)$ such that $j_*\gamma = \sigma$. But then $i_* : H_0(A) \to H_0(X)$ cannot be surjective: if you have $\gamma = i_*\alpha$, then $j_*\gamma = j_*i_*\alpha = 0$ by exactness.
  • Or $\partial \sigma$ is nonzero. But then $i_*(\partial\sigma) = 0$ by exactness, so $i_* : H_0(A) \to H_0(X)$ is not injective.

So all that's left to prove is that $i_* H_0(A) \to H_0(X)$ is not injective iff there is a path component of $X$ containing at least two path components of $A$.

Again, one direction is clear. Suppose a path component of $X$ contains two path components of $A$. Pick points $a,b \in A$ in the two distinct path components. Then $[a]-[b] \in H_0(A)$ is nonzero, but $i_*[a] = i_*[b]$ thus $i_*([a]-[b]) = 0$.

Conversely suppose that $i_* : H_0(A) \to H_0(X)$ is not injective. Let $0 \neq \alpha \in H_0(A)$ be such that $i_*\alpha = 0$. Write $\alpha = \sum_{k \in I} n_k [a_k]$ as a sum of vertices. Moreover assume that all the $a_i$ are in different path components.

Of course, $i_*(\alpha) = \sum_{k \in I} n_k [a_i]$ viewed as $0$-cycles in $H_0(X)$. If the path components corresponding to the $a_i$ were all different in $X$ then this wouldn't be possible unless $n_i = 0$ for all $i$ (here you use the fact that $H_0(X)$ is the free abelian group on the path components of $X$). But $\alpha \neq 0$, so this isn't possible. It follows that at least two of the $a_i$ are in the same path component in $X$.

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