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

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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.

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(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 H1(A) \xrightarrow{i} H1(X) \xrightarrow{j} H1(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.