# 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.

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Teri Hatcher? – miracle173 Nov 10 '12 at 7:04
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

Let $a,b \in A$ lie in separate path components. So that $[a],[b]$ represent distinct classes in $H_0(A)$. If $i(a),i(b)$ lie in the same path component in $X$ then there is a path, or equivalently a 1-simplex $\sigma: [0,1] \rightarrow X$ with $\sigma(0)=i(a)$ and $\sigma(1)=i(b)$. Thus $\partial \sigma = i(a) - i(b)$ which shows that $i(a)$ and $i(b)$ represent the same class in $H_0(X)$. Thus $i_* :H_0(A) \rightarrow H_0(A)$ is not injective.
That gives one direction. I don't think the opposite direction should present too many problems, if you look a the above argument and see how it changes when $i(a)$ and $i(b)$ lie in different path components.