Theorem: Suppose $A$ is a retract of $X$, $r: X \rightarrow A$ is the retraction, and $i : A \rightarrow X$ is the inclusion. As usual, let $i_*$ and $r_*$ be the induced homomorphisms that operate on paths. If $i_*\pi(A) = \im(i_*)$ is a normal subgroup of $\pi(X)$, then $\pi(X)$ is the direct product of $\im(i_*)$ and $\ker(r_*)$.

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Proof: First, note that $\ker(r_*)$ is necessarily a normal subgroup of $\pi(X)$ ($r_*$ is a homomorphism so its kernel is a normal subgroup). The fact that $\im(i_*)$ is normal is a hypothesis.

Second, note that $\ker(r_*) \cap \im(i_*) = 1$. This is apparent from noting that any non-trivial element in $\im(i_*)$ would necessarily not be in $\ker(r_*)$.

Now, we claim that $(\ker(r_*))(\im(i_*)) = \pi(X)$. Clearly, $\subseteq$ is trivial, so we show $\supseteq$. Consider an arbitrary $\alpha \in \pi(X)$. Let

$$\beta = i_* r_* (\alpha) \qquad \gamma = i_* r_* (\alpha)^{-1} \alpha$$

Note that $\beta \in \im(i_*)$, trivially. To show that $\gamma \in \ker(r_*)$, we instead show that $\gamma^{-1} \in \ker(r_*)$. Note that:

$$\begin{align*} r_* (\gamma^{-1}) &= r_* (\alpha^{-1} i_* r_* (\alpha)) \\ &= r_* (\alpha^{-1}) r_* (i_* r_* (\alpha)) \\ &= r_* (\alpha)^{-1} r_* i_* r_* (\alpha) \\ &= r_*(\alpha)^{-1} r_* (\alpha) \\ &= 1 \end{align*}$$

And so, $\gamma^{-1} \in \ker(r_*)$ and $\gamma$ is as well. Then, simply observe that $\beta \gamma = \alpha$. So, any element of $\pi(X)$ can be written as a product from $\ker(r_*)$ and $\im(i_*)$, and we've shown the claim.

So, we have two normal subgroups whose intersection is trivial, and whose product is all of $\pi(X)$. Therefore, we have $\pi(X) = \ker(r_*) \times \im(i_*)$.