1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

enter image description here

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

enter image description here

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.