# A relative version of Urysohn's Lemma?

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 $s(X)$?

• 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). – Włodzimierz Holsztyński Mar 26 '15 at 17:24
• Why should normality imply the second property? – user1688 Mar 26 '15 at 19:12
• 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?). – Włodzimierz Holsztyński Mar 27 '15 at 6:29
• Oh, you mean that this is a single question, not two independent questions. Sorry. – Włodzimierz Holsztyński Mar 27 '15 at 6:31