1
$\begingroup$

I have the following setup:

$X, Y$ are topological spaces (if it helps, they can both be $T_1$ and normal. They can even be countably paracompact. They can't be assumed paracompact). $V$ is a normed space (it can be Banach if you like). $f : X \to Y$ is a perfect surjection.

I have continuous and bounded $g : X \to V$ and given $\epsilon > 0$ would like to find continuous $h : Y \to V$ such that $d(h(x), g(f^{-1}(x))) < \epsilon$

Is there some sort of selection theorem that will let me do this? I've used the Michael selection theorem to good effect elsewhere, but it doesn't apply here due to the lack of convexity of the target sets (even if they were convex the hypotheses don't apply due to potential non-paracompactness of Y, but one might be able to work something out using countable paracompactness and compactness of the targets).

$\endgroup$
3
  • 1
    $\begingroup$ Let $X=Y=S^1$, $f$ be a standard 2-fold covering, $V=\mathbb R^2$, $g$ be a standard embedding of the circle into the plane. If I understand your notation correctly, there is no $h$ if $\epsilon$ is small. $\endgroup$ Apr 7, 2010 at 9:40
  • 1
    $\begingroup$ By the way, if you rewrite that comment as an answer then I can accept it :-) $\endgroup$ Apr 7, 2010 at 11:12
  • $\begingroup$ Ok, I will do so. $\endgroup$ Apr 7, 2010 at 17:57

1 Answer 1

4
$\begingroup$

Consider $X=Y=S^1$. Let $f:X\to Y$ be a 2-fold covering and $g:X\to\mathbb R^2$ the standard embedding (whose image is a unit circle). Assume $\epsilon<1$, then there is no map $h$ with the desired property.

Indeed, if $d(h(x),g(f^{-1}(x)))<\epsilon$, then there is a unique $y\in f^{-1}(x)$ such that $d(h(x),g(y))<\epsilon$. Obviously $y=:u(x)$ depends continuously on $x$. Thus we obtain a continuous map $u:Y\to X$ such that $f\circ u=id_Y$. Such a map does not exist because $f$ (and hence any map of the form $f\circ u$) induces a non-surjective homomorphism of fundamental groups.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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