Path-components and relative homology

Question

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?

Answer 1

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 $A = \{ 0,1 \}$.

The proof of (b) is correct.