Diagram which commutes up to homotopy commutes strictly

Question

Let $M$ and $N$ be manifolds of same dimension with boundary. Let $f \colon M \to N$ be a continuous map. Apparently if the diagram $\require{AMScd}$ \begin{CD} \partial M @>>>M\\ @VVV @VVV{f}\\ \partial N @>>> N \end{CD}

commutes up to homotopy, then there is a map $\tilde f \colon M \to N$ homotopic to $f$ such that the diagram commutes strict. I do not see why this is true.

Answer 1

This is because $\partial M \rightarrow M$ is a Hurewicz cofibration, and if $i:A \rightarrow X$ is a Hurewicz cofibration then if you have a diagram

which commutes up to homotopy, then you can replace $f$ by a homotopic map $\gamma$ so that the diagram

strictly commutes.

This is problem 5.3 in Jeffrey Stroms "Modern classical homotopy theory" found on page 101.

To prove that $\partial M \rightarrow M $ is a Hurewicz cofibration have a look at this math overflow answer and note that all Serre cofibrations are also Hurewicz cofibrations.