# Diagram which commutes up to homotopy commutes strictly

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.

## 1 Answer

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.