7
$\begingroup$

Say that a space (= compact metrizable space) $X$ is locally strictly contractible if, for every $p\in X$ and neighborhood $U$ of $p$, there is a neighborhood $V$ of $p$ which can be contracted to $p$ within $U$ relative to $p$, i.e. by a homotopy leaving $p$ fixed. (This may be bad terminology, and I welcome alternatives.)

Every ANR is locally strictly contractible. What about the converse? Borsuk's example of a locally contractible space which is not an ANR is not locally strictly contractible.

Incidentally, "locally contractible" is defined in different ways in various references. What is the currently accepted definition?

Improve this question
$\endgroup$
20
  • 1
    $\begingroup$ Bruce: beware that 'strictly contractible' does have another meaning in the literature, not unrelated to yours: sciencedirect.com/science/article/pii/S0166864101000086 $\endgroup$
    – Sergey Melikhov
    Commented Jul 20, 2012 at 18:08
  • 1
    $\begingroup$ Upon some reflection, Borsuk's example does actually seem to be, hmm, pointed locally contractible. It's an interesting question though if there exists a locally contractible compactum that is not pointed locally contractible. It would have to be infinite-dimensional. As for Borsuk's example, it is not locally equi-connected in the sense of Fox and Dugundgi (or equivalently, not uniformly locally contractible in the sense of Isbell). A long-standing open problem which has considerable literature is whether local equi-connectedness is equivalent to the ANR property for compacta. $\endgroup$
    – Sergey Melikhov
    Commented Jul 20, 2012 at 23:06
  • 1
    $\begingroup$ I guess I still don't see how to contract Borsuk's example to, say, the point with $x_1=0$ and $x_n=1/2n$ for $n>1$. It seems that any contraction of a neighborhood has to send this point via a point with $n$'th coordinate 0 or $1/n$ for some $n>1$. $\endgroup$
    – Bruce Blackadar
    Commented Jul 21, 2012 at 14:06
  • 1
    $\begingroup$ The problem is that to contract a neighborhood, you have to be able to work in a coordinate $m$ for which the neighborhood doesn't hit both ends of $X_m$, so points can be moved around one end. But for any neighborhood of the point $x$, there are only finitely many such coordinates to work with, so the end you move around can't be passed off to infinity. (Hope this cryptic description makes sense.) $\endgroup$
    – Bruce Blackadar
    Commented Jul 21, 2012 at 16:44
  • 1
    $\begingroup$ Bruce, I now see that you were right. The motion I've described is of the order of $1/i$ but it affects points that about $(1/m)$-close to $x$. Here $m$ could be large, so for the homotopy to be continuous, $x$ has to move as well. Sorry for taking your time, and thank you for explaining Borsuk's example, which as you see I didn't quite understand. $\endgroup$
    – Sergey Melikhov
    Commented Jul 22, 2012 at 21:37

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .