Show that a retract of a cofibration is also a cofibration.

Question

Here is the question:

Suppose that $g: A \rightarrow C $ is a retract of $f: B \rightarrow D.$ Show that if $f$ is a cofibration, then so is $g.$ Could anyone help me in answering this question, please?

Answer 1

HINT:

$f:B \rightarrow D$ is a cofibration. It satisfies the homotopy extension property with respect to any map $\alpha:D \rightarrow Y$.

$g:A \rightarrow C$ is a retract of $f$. There is a particular commutative diagram you can draw defining this property.

You now wish to show that $g$ is a cofibration. You need it to satisfy the homotopy extension property for any map $\hat{\alpha}:C \rightarrow Z$.

Suppose you're given a homotopy $H:A \times I\rightarrow Z$ satisfying the appropriate properties. We get a map $\hat{\alpha} \circ s:D \rightarrow Z$ and a homotopy $H \circ (r \times id): B \times I \rightarrow Z$ by precomposing with the retraction $r: B \rightarrow A$ and $s:D \rightarrow C$.

Since $f$ is a cofibration, this homotopy can be extended to a homotopy $D \times I \rightarrow Z$.

Now precompose with the inclusion to get a homotopy $C \times I \rightarrow Z$, and check everything commutes, and you're done.