I'm proving the following lemma (this is part of a proof of Hurewicz's theorem)

Lemma. Let $n\geq1$. For $n> 1$ assume known that $\pi_{n−1}(S^{ n−1}, z)$ is free abelian of rank $1$. Let $f$ be a continuous self map of $(D^n,\partial D^n)$ of degree $\pm1$. Then $f$ is pair-homotopic to the identity if $\deg(f) = 1$

At one point of the proof, we proved the existence of an homotopy $H:S^{n-1}\times I\rightarrow S^{n-1}$ between $f|_{S^{n-1}}$ and the identity $\text{id}$ of $S^{n-1}$. Then, the professor says that

[The homotopy $H$] induces a map $D^n\times\{0\}\cup S^{n-1}\times I\cup D^n\times\{1\}\xrightarrow{f|_{S^{n-1}}\cup H\cup\text{id}}D^n$. Since $D^n\times I$ can be obtained from $D^n\times\{0\}\cup S^{n-1}\times I\cup D^n\times\{1\}$ by attaching an $(n + 1)$-cell and $D^n$ is contractible, there is a continuous extension $\bar H: D^n \times I\rightarrow D^n$

I don't understand why we need $D^n$ to be contractable. Moreover, I would say that the homotopy $\bar H$ can be described explicitly: First, $D^n\times\{0\}\cup S^{n-1}\times I$ is a retraction of $D^n\times I$, so take $$p:D^n\times I\rightarrow D^n\times\{0\}\cup S^{n-1}\times I$$ identity on the subspace, then $$\tilde H:D^n\times I\xrightarrow{p}D^n\times\{0\}\cup S^{n-1}\times I\xrightarrow{f|_{S^{n-1}}\cup H}S^n\hookrightarrow D^n$$is an homotopy between $f$ and a map $g:D^n\rightarrow D^n$ which is the identity on $S^{n-1}$. Then consider the map $$\hat H:D^n\times I\rightarrow D^n,\qquad\hat H(v,t)=g(v)(1-t)+vt$$Then $\hat H$ is a pair-homotopy between $g$ and the identity of $D^n$, hence we can juxtapose $\tilde H,\hat H$ to get a pair-homotopy between $f$ and the identity.

Is there something wrong with my argument? Any feedback is appreciated.


1 Answer 1


Your argument is correct. But let me first note that you need $f\cup H\cup\text{id}$ instead of $f|_{S^{n-1}}\cup H\cup\text{id}$.

The space $B = D^n\times\{0\}\cup S^{n-1}\times I\cup D^n\times\{1\}$ is the boundary of the $(n+1)$-cell $D^n \times I \approx D^{n+1}$ and you have a map $\phi : B \to X$ with $X = D^n$. It is well-known that $\phi$ has a continuous extension $\bar H : D^n \times I \to X$ if and only if $\phi$ is null-homotopic. If $X$ is contractible, this is automatically true. Hence for $X = D^n$ we are finished. Only if you do not know that $X$ is contractible, you must in fact explicitly construct $\bar H$. But you can do it also in the present case, and your approach works.

  • $\begingroup$ Of course, I didn't think about this condition for the existence of an extension. Thanks $\endgroup$
    – Alessandro
    Jan 13, 2022 at 7:10
  • $\begingroup$ @Alessandro By the way, your proof uses the convexity of $D^n$ which is much stronger than contractibility. $\endgroup$
    – Paul Frost
    Jan 13, 2022 at 10:01

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