Problem: $Y \subset X$ be a subcomplex of a C. W. complex. Suppose that there is a retraction $r: X \to Y$. Prove that there is a map from $\Sigma X \to \Sigma Y \vee \Sigma(X/Y)$ which is a homology isomorphism.
Taking the example of $X$ being the annulus and $Y$ the inner circle I get that the given spaces of suspensions are homotopy equivalent.
In general I think we need to show that $\Sigma X $ and $ \Sigma Y \vee \Sigma(X/Y)$ are homotopy equivalent. A homotopy equivalence, say $f$, will clearly give isomorphisms on all the homology groups. But I do not know how to show the homotopy equivalence. In fact it is not necessary I think since the question only asks for homology isomorphisms.
Using the hint from @ConnorMalin.
For any $X \in \mathcal{Top}_*$ $[\sum{X}, Z]$ has a natural group structure(denote by +) for any $Z$. The homology functors are linear wrt. to this group structure. $\cdots (1)$.
From the axioms of homology, we have $\cdots \to H_n(Y) \to H_n(X) \to H_n(X, Y) \to H_{n-1}(A) \to \cdots$. Since $(X, Y)$ is a good pair we get $\cdots \to \tilde{H}_n(Y) \to \tilde{H}_n(X) \to \tilde{H}_n(X/Y) \to \tilde{H}_{n-1}(A) \to \cdots$
$r : X \to Y$ being a retraction gives that the boundary maps are zero and the resulting short exact sequences split. We get,
$ \tilde{H}_n(X) \xrightarrow{[r^* q^*]^T} \tilde{H}_n(Y) \oplus \tilde{H}_n(X/Y)$.
Using another axiom for reduced homology that says $\tilde{H}_n$ is naturally isomorphic to $\tilde{H}_{n+1}\circ \sum$ we get
$ \tilde{H}_{n+1} (\sum X) \xrightarrow{[\sum r^* \sum q^*]^T} \tilde{H}_n(\sum Y) \oplus \tilde{H}_n(\sum X/Y)$.
Also, the inclusions $i_Y$ and $i_{X/Y}$ of $\sum X$ and $\sum Y$ respectively in $\sum X \vee \sum X/Y$ induce the isomorphism
$\tilde{H}_n(\sum Y) \oplus \tilde{H}_n(\sum X/Y) \xrightarrow{[i_Y^* i_{X/Y}^{*}]} \tilde{H}_n (\sum Y \vee \sum X/Y)$.
By $(1)$ the composition of the previous two isomorphisms is induced by $i_y \circ \sum r + i_{X/Y} \circ \sum q$