3
$\begingroup$

A fundamental result in topology is that the $n$-sphere is not a retract of the $n+1$-ball. It implies that the $n$-sphere is not an absolute retract.

Is there a generalization from the sphere to closed manifolds (compact manifolds without boundary)? It would be the statement that no closed manifold is an absolute retract.

Improve this question
$\endgroup$
1
  • $\begingroup$ Perhaps a bit more properties need to be assumed about the manifold? Do you know any sufficient conditions for your theorems to work? $\endgroup$
    – Sam Sanders
    Commented Mar 15, 2021 at 11:38

1 Answer 1

Reset to default
12
$\begingroup$

A metrizable space is an absolute retract (AR) if and only if it is an absolute neighbourhood retract (ANR) and it is contractible.

Closed manifolds are not contractible (if $\dim M=n$ look at $H_n(M)$) hence they are not ARs. Note however that contractibility is the only obstruction for manifolds, since every topological manifold is an ANR.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you! I see here that $H_n(M; \mathbb{Z}_2)=\mathbb{Z}_2$. $\endgroup$
    – mathieu
    Commented Mar 15, 2021 at 12:07
  • 1
    $\begingroup$ @mathieu: Yes, every closed topological manifold is $\mathbb{Z}/2\mathbb{Z}$-orientable. For example, you can find the result and its proof in Hatcher's book, p.235 in 2nd ed. $\endgroup$
    – M.G.
    Commented Mar 15, 2021 at 17:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .