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The Borsuk-Ulam theorem is equivalent to $S^{n-1}$ not being a retract of $B^n$. <totally wrong! or else 2+2=4 is equivalent to the Poincare conjecture/thm>

How shall i prove the following stronger version :

There is no continuous map $f:\Delta_n \to \partial \Delta_n$ such that every face is mapped to itself. <the Borsuk's homotopy extension thm since $f|S^{n-1}$ would be homotopic to the identity on $\ S^{n-1}$>

In this case, Tucker's lemma comes to mind, as it does say something similar.
Any thoughts?

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  • $\begingroup$ If $f:\Delta_n\to \Delta_n$ maps every face $F$ to itself, one can consider the topological degree of $f_{|\partial F}:\partial F\to\partial F$ and prove by induction on dim(F) that it is $1$. So $f(F)=F$ for all faces, and in particular $f(\Delta_n)=\Delta_n$. $\endgroup$
    – Pietro Majer
    Commented Jul 27, 2020 at 20:39
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    $\begingroup$ Does this answer your question? Map from simplex to itself that preserves sub-simplices $\endgroup$
    – Moritz Firsching
    Commented Jul 27, 2020 at 21:05
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    $\begingroup$ @Wlod AA, I'm pretty sure that way to use edits is against the rules here. If you have something to say, try comments or answers... $\endgroup$
    – Achim Krause
    Commented Jul 27, 2020 at 22:46

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