# retract, inclusion, normal subgroup

Prove that, if $A$ is a retract of $X$, $r:X \rightarrow A$ is a retraction, $i: A \rightarrow X$ is the inclusion, and $i_*\pi(A)$ is a normal subgroup of $\pi(X)$, then $\pi(X)$ is the direct product of the subgroups image $i_*$ and kernel $r_*$.
[So I need to show $\pi_1(X) = \text{ker}(r_*) \times i_*(\pi_1(A)).$]