5
$\begingroup$

A continuous map $d:X\to A$ is called domination if there exists a map $u:A\to X$ so that $d\circ u\simeq 1_A$.

Is there a domination map $d:P\to P$ of a finite polyhedron $P$ so that $d$ is not a homotopy equivalence?

$\endgroup$

1 Answer 1

3
$\begingroup$

This question apparently goes back to Karol Borsuk, at least in spirit. An interesting discussion together with a history of the problem can be found in a paper of Danuta Kołodziejczyk, Polyhedra for which every homotopy domination over itself is a homotopy equivalence, arxiv:1411.1032.

For instance, if $P$ is a manifold [Bernstein-Ganea, 1959] or, more generally, a Poincaré complex [Kwasik, 1984], then every domination of $P$ is a homotopy equivalence. The same is true for polyhedra $P$ with polycyclic-by-finite fundamental group [Kołodziejczyk, 2005].

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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