Massey Exercise 2.4.3

Theorem: Suppose is a retract of , is the retraction, and is the inclusion. As usual, let and be the induced homomorphisms that operate on paths. If is a normal subgroup of , then is the direct product of and .

Proof: First, note that is necessarily a normal subgroup of ( is a homomorphism so its kernel is a normal subgroup). The fact that is normal is a hypothesis.

Second, note that . This is apparent from noting that any non-trivial element in would necessarily not be in .

Now, we claim that . Clearly, is trivial, so we show . Consider an arbitrary . Let

Note that , trivially. To show that , we instead show that . Note that:

And so, and is as well. Then, simply observe that . So, any element of can be written as a product from and , and we’ve shown the claim.

So, we have two normal subgroups whose intersection is trivial, and whose product is all of . Therefore, we have .