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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).

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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. – Sergei Ivanov Apr 7 '10 at 9:40
Hm. Quite right. Thanks. – David R. MacIver Apr 7 '10 at 10:03
By the way, if you rewrite that comment as an answer then I can accept it :-) – David R. MacIver Apr 7 '10 at 11:12
Ok, I will do so. – Sergei Ivanov Apr 7 '10 at 17:57
up vote 4 down vote accepted

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.

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