Approximate selection theorems for factoring through perfect maps - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T00:01:13Z http://mathoverflow.net/feeds/question/20608 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20608/approximate-selection-theorems-for-factoring-through-perfect-maps Approximate selection theorems for factoring through perfect maps David R. MacIver 2010-04-07T08:20:45Z 2010-04-07T18:30:01Z <p>I have the following setup:</p> <p>$X, Y$ are topological spaces (if it helps, they can both be $T_1$ and normal. They can even be countably paracompact. They <em>can't</em> be assumed paracompact). $V$ is a normed space (it can be Banach if you like). $f : X \to Y$ is a perfect surjection.</p> <p>I have continuous and bounded $g : X \to V$ and given $\epsilon > 0$ would like to find continuous $h : Y \to V$ such that $d(h(x), g(f^{-1}(x))) &lt; \epsilon$</p> <p>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).</p> http://mathoverflow.net/questions/20608/approximate-selection-theorems-for-factoring-through-perfect-maps/20653#20653 Answer by Sergei Ivanov for Approximate selection theorems for factoring through perfect maps Sergei Ivanov 2010-04-07T18:30:01Z 2010-04-07T18:30:01Z <p>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 $\epsilon&lt;1$, then there is no map $h$ with the desired property.</p> <p>Indeed, if $d(h(x),g(f^{-1}(x)))&lt;\epsilon$, then there is a unique $y\in f^{-1}(x)$ such that $d(h(x),g(y))&lt;\epsilon$. 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.</p>