If $A$ is a retract of $X$ (via $r:X\to A$), then $H_q(X)\cong \text{im }i_*\oplus \ker r_*$

Question

I need to prove the following statement:

If $A$ is a retract of $X$ (via $r:X\to A$), then $H_q(X)\cong \text{im }i_*\oplus \ker r_*$

It is well know that $r\circ i=id_A$ implies that $r_*$ and $i_*$ are surjective and injective respectively. From here, is it possible conclude that $H_q(X)\cong \text{im }i_*\oplus \ker r_*$?

Answer 1

We have the following sequence.

$$0 \to \ker r_* \to H_q(X) \stackrel{r_*}\to H_q(A) \to 0 ,$$ which is exact since $r$ is surjective as you said.

It is also split, with splitting given by $i_*$, since $r_* \circ i_*=\mathrm{Id}$ by functoriality.

Therefore, $H_q(X) \cong H_q(A) \oplus \ker r_*$. But $i_*$ is injective, therefore an isomorphim with its image, and the result follows.