# Hatcher Problem 2.1.16 (b)

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?

Thanks.

• 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 $H_1(A)\to H_1(X)$. – Stefan Hamcke Nov 10 '12 at 17:41
• @StefanH. I have corrected the first point in the comment. I don't understand which map you're talking about in the second. – user38268 Nov 10 '12 at 22:40
• Why must $\sigma$ be a loop? Are you assuming that $\tau$ is a single vertex rather than an arbitrary 0-chain in $A$? – John Palmieri Nov 10 '12 at 23:49
• @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 $X$. – Stefan Hamcke Nov 11 '12 at 13:06
• @StefanH. I have corrected it now. Thanks. – 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 $$X$$.