Let $f:Y\to X$ be a continuous surjective map between locally compact Hausdorff spaces. Assume there is a continuous section $s:X\to Y$ which has closed image and is a homeomorphism to the image. I have two questions: First, is it true that there exists a neighborhood $U$ of $s(X)$ such that $U\cap s^{-1}(x)$ is compact for every $x\in X$. Second, does there exist a continuous function $f:Y\to [0,1]$ such that $f$ is supported inside $U$ and $f$ is constant equal to 1 on $s(X)$?
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$\begingroup$ The second question is instantly true when $\ Y\ $ is normal (right?), so that it is non-trivial--possibly difficult--for non-normal Y (which feel a bit weird). $\endgroup$– Włodzimierz HolsztyńskiMar 26, 2015 at 17:24
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$\begingroup$ Why should normality imply the second property? $\endgroup$– user1688Mar 26, 2015 at 19:12
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$\begingroup$ When $Y$ is normal, and $s(X)$ is closed in $Y$ then there kis a Urysohn function $f:Y\rightarrow[0;1]$ such that $f|s(X)=1$, and $f|Y\setminus U=0$ (am I missing something?). $\endgroup$– Włodzimierz HolsztyńskiMar 27, 2015 at 6:29
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$\begingroup$ Oh, you mean that this is a single question, not two independent questions. Sorry. $\endgroup$– Włodzimierz HolsztyńskiMar 27, 2015 at 6:31
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