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If $X = A \cup B$ is a disconnection then $H_n(X) \cong H_n(A) \oplus H_n(B)$

If I use the Mayer-Vietoris sequence I can prove this easily, however the book I'm reading hinted at using excision to prove this.

So I tried to prove this using excision but I ended up getting stuck, this is what I've done so far.

Since $B \subseteq X$ we obtain a long exact sequence

$$ \dots \to H_n(B) \to H_n(X) \to H_n(X, B) \to H_{n-1}(B) \to \dots$$

and observe that because $X = A \cup B$ by Excision we get $H_n(A) \cong H_n(X, B)$ and so we obtain the following long exact sequence

$$ \dots \to H_n(B) \xrightarrow{f} H_n(X) \xrightarrow{g} H_n(A) \to H_{n-1}(B) \to \dots$$

This is where I get stuck, because if $f$ was injective and $g$ was surjective I could use exactness and the first isomorphism theorem to get the desired result, however I have no way of knowing if $f$ is injective and if $g$ is surjective.

I'm not sure how to proceed and prove this using excision alone.

Perturbative
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    You could always argue by hand that $f$ and $g$ are injective and surjective chain $\sigma$ in $X$. Then you can write $\sigma=\sigma_A+\sigma_B$ where $\sigma_A$ is the sum of simplices lying in $A$ and so forth. It is pretty clear that $\partial \sigma_B=0$. So your original cycle is actually trivial just in $A$. However this line of argument can be readily adapted to not even use excision. The chain complex $C_n(X)$ itself breaks up as a direct sum $C_n(A)\oplus C_n(B)$. – Cheerful Parsnip Dec 05 '18 at 20:53
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    I think any proof that doesn't just check this at the chain level is silly. –  Dec 05 '18 at 21:39
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    Note that if $X$ is the disjoint union of $A\sqcup B$, then both inclusions admit retractions and therefore the inclusion has to induce an injective map on homology and the retraction, which induces the map from $H_(X,B)\to H_(A)$ given via excision has to be surjective and therefore the long exact sequence breaks down to a short exact sequence, which admits a splitting. – ThorbenK Dec 05 '18 at 22:39

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Let $H_*$ be any homology theory satisfying the Eilenberg-Steenrod axioms with the possible exception of the dimension axiom. We shall prove that $H_n(X) \cong H_n(A) \oplus H_n(B)$.

Let $i : A \to X$ denote the inclusion map. Choose $a \in A$ and define $r : X \to A$ by $r(x) = x$ for $x \in A$ and $r(x) = a$ for $x \in B$. Then $ri = id$, hence $i_* : H_n(A) \to H_n(X)$ is a monomorphism with left inverse $r_*$. This shows that the long exact sequence of the pair $(X,A)$ collapses to split short exact sequences $$0 \to H_n(A) \stackrel{i_*}{\rightarrow} H_n(X) \stackrel{j_*}{\rightarrow} H_n(X,A) \to 0 .$$ We conclude $$H_n(X) \approx H_n(A) \oplus H_n(X,A) .$$ Excision gives us an isomorphism $H_n(B) = H_n(B,\emptyset) \to H_n(X,A)$ which completes the proof.

Paul Frost
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