Retract implies derived subgroup equals intersection with whole derived subgroup

Statement with symbols
Suppose $$H$$ is a retract of a group $$G$$: in other words, $$H$$ is a subgroup of $$G$$ such that there is a retraction from $$G$$ to $$H$$: a surjective homomorphism $$\alpha:G \to H$$ such that the restriction of $$\alpha$$ to $$H$$ is the identity map. Then, $$H \cap [G,G] = [H,H]$$.

Proof
Given: A group $$G$$, a subgroup $$H$$ with retraction $$\alpha:G \to H$$.

To prove: $$H \cap [G,G] = [H,H]$$.

Proof: By definition, we have $$[H,H] \le H$$ and $$[H,H] \le [G,G]$$. Thus, we have $$[H,H] \le H \cap [G,G]$$. We need to prove the reverse inclusion.

Note that a generic element of $$[G,G]$$ is of the form:

$$g = [a_1,b_1][a_1,b_2] \dots [a_n,b_n]$$

where $$a_i,b_i \in G$$. Note that we do not need negative powers, since the inverse of a commutator is again a commutator.

Since $$\alpha$$ is a homomorphism, we have:

$$\alpha (g) = \alpha([a_1,b_1][a_2,b_2] \dots [a_n,b_n]) = [\alpha(a_1),\alpha(b_1)][\alpha(a_2),\alpha(b_2)] \dots [\alpha(a_n),\alpha(b_n)]$$.

We want to show that if $$g \in H$$, then $$g \in [H,H]$$.

Suppose $$g \in H$$. Then, since $$\alpha$$ is a retraction, $$\alpha(g) = g$$, so we get that $$g$$ is a product of commutators $$[\alpha(a_i),\alpha(b_i)]$$ between elements of $$H$$. Thus, $$g \in [H,H]$$.