# Path-components and relative homology

This is an exercise from Algebraic Topology book of Hatcher:

Exercise 2.1.16, pg. 132:

$$(a)$$ Show that $$H_0(X;A) = 0$$ if $$A$$ meets each path-component of $$X$$.

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

The $$(a)$$ part of this question is extensively discussed on $$H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall$$ path-components $$X_i$$. Also duplicates of this question are $$H_0(X,A) = 0 \iff A$$ meets each path-component of $$X$$. and For any Subspace $$A$$ of a Path-Connected Space $$X$$, we have $$H_0(X, A)=0$$.

First, I offer a short and somehow alternative proof for $$(a)$$:

For the pair $$(X, A)$$, we have a long exact sequence: $$\ldots \to H_1(X,A) \to H_0(A) \xrightarrow{f} H_0(X) \to H_0(X,A) \to 0$$ By Proposition 2.7 of Hatcher, we know that for any space $$X$$, $$H_0(X)$$ is a direct sum of $$\mathbb{Z}$$’s, one for each path-component of $$X$$.

If $$H_0(X, A)=0$$, then $$f$$ is surjective so that the number of path components in $$A$$ is equal to the number of path components in $$X$$. Therefore, $$A$$ meets each path-component of $$X$$.

Conversely, suppose $$A$$ meets each path-component of $$X$$. Then clearly there is a one-to-one correspondence between the path components of $$A$$ and path-components of $$X$$, which implies that $$f$$ is surjective. Therefore, $$H_0(X, A) = 0$$.

Is there any gap in the proof?

The $$(b)$$ part of this question is also discussed on $$H_1(X,A) = 0 \iff H_1(A) \rightarrow H_1(X)$$ surjective and $$X_i$$ contains no more than one path-component of $$A$$.

But there is again no need to write long proof as the suggestion of @shaun-ault.

For the pair $$(X, A)$$, consider the long exact sequence $$\ldots \to H_1(A) \xrightarrow{g} H_1(X) \to H_1(X, A) \to H_0(A) \xrightarrow{f} H_0(X) \to H_0(X,A) \to 0$$

If $$H_1(X, A) = 0$$, then $$g$$ is surjective and $$f$$ is injective which implies that each path-component of $$X$$ contains at most one path-component of $$A$$.

Conversely, suppose that $$g$$ is surjective and each path-component of $$X$$ contains at most one path-component of $$A$$. The latter assumption implies that $$f$$ is injective. Together with the former assumption, we conclude that $$H_1(X, A) = 0$$.

Again, is there any gap in the second part of the proof?

Your proof of (a) is a little imprecise when you state "If $$H_0(X, A)=0$$, then $$f$$ is surjective so that the number of path components in $$A$$ is equal to the number of path components in $$X$$."
In fact, $$f$$ being surjective means that for each path component $$X_i$$ of $$X$$ there exists at least one path component of $$A_j$$ of $$A$$ such that $$A_j \subset X_i$$. In general this will not be a one-to-one correspondence between the path components of $$A$$ and the path-components of $$X$$.
As an example consider $$X = [0,1]$$ and $$A = \{ 0,1 \}$$.