Massey Exercise 2.4.3 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_*). 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: % 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_*).