(a) Theorem: $H_0(X,A) = 0$ if and only if $A$ meets each path component of $X$.

---

Proof: We start with a lemma, and then apply Proposition 2.6 from Hatcher.

Lemma: For $A \subset X$ with $X$ path-connected, $H_0(X,A)$ is trivial.

Proof: From the long exact sequence relating homology groups and relative homology groups, we have:

$$\dots \rightarrow H_0(A) \xrightarrow{i_*} H_0(X) \xrightarrow{j_*} H_0(X,A) \rightarrow 0$$

$H_0(X)$ is generated by a single basepoint – in particular, by some basepoint $a \in A$. Thus, $H_0(X)$ is generated by $a$, and $a$ is at least among the generators of $H_0(A)$ ($A$ may not be path connected itself). Then, we must at least have that $i_*$ is surjective. But this means $\ker j_*$ is $H_0(X)$, and $\im j_* = \set{0} = H_0(X,A)$.

$$\mjqed$$

Now, we use the fact that for a general space $X$, there is an isomorphism $H_n(X) \cong \bigoplus_i H_n(X_i)$ for each path component of $X$. Let $A_i = A \cap X_i$ be the non-empty intersection of $A$ with each path component of $X$. Then, we have:

$$\bigoplus_i H_0(A_i) \xrightarrow{i_*} \bigoplus_i H_0(X_i) \xrightarrow{j_*} \bigoplus_i H_0(X_i, A_i) \rightarrow 0$$

The boundary maps in this sequence all split across the direct sums, and we apply the lemma (note each $X_i$ is path-connected) to each component, yielding $H_0(X_i, A_i) = 0$ for all $i$. Then, $\bigoplus_i H_0(X_i, A_i) = H_0(X,A) = 0$ as well.

---

(b) Theorem: $H_1(X,A) = 0$ if and only if $H_1(A) \rightarrow H_1(X)$ is surjective and each path-component of $X$ contains at most one path component of $A$.

---

Proof: Again, start with a lemma:

Lemma: For $A \subset X$ with $X$ path-connected, $H_1(X,A)$ is trivial if and only if $i_* : H_1(A) \rightarrow H_1(X)$ is surjective and $A$ is path-connected.

Proof: Consider the relevant portion of the long exact sequence:

$$\dots \rightarrow H_1(A) \xrightarrow{i_*} H_1(X) \xrightarrow{j_*} H_1(X,A) \xrightarrow{f} H_0(A) \xrightarrow{i_*} H_0(X) \rightarrow \dots$$

From a previous result, we know that $H_1(X,A)$ is trivial if and only if $i_* : H_1(A) \rightarrow H_1(X)$ is surjective (a hypothesis), and $i_* : H_0(A) \rightarrow H_0(X)$ is injective. We show the latter.

Note that $H_0(X)$ is generated by a single basepoint, in particular a basepoint in $A$. Then, $H_0(A)$ is generated by the same point (here we use that $A$ is path-connected). Thus, $H_0(A) = H_0(X)$ in fact, and $i_*$ is clearly injective.

$$\mjqed$$

Consider now a general space $X$ with path components $\set{X_i}$, and let $A_i = A \cap X_i$. Then, again we consider the exact sequence:

$$\bigoplus_i H_1(A_i) \rightarrow \bigoplus_i H_1(X_i) \rightarrow \bigoplus_i H_1(X_i, A_i) \rightarrow \bigoplus_i H_0(A_i) \rightarrow \bigoplus_i H_0(X_i)$$

The boundary maps split across the direct sums, and we apply the lemma to each component (this time, note that each $A_i$ is path-connected because each path component of $X$ contains no more than one path component of $A$). Then, we conclude $H_1(X_i, A_i)$ is trivial for all $i$, and thus $H_1(X,A)$ is trivial.