Let $f: A \to Y $ be a continuous function, where $A$ is a closed subset of a space $X$, then is it true that $f$ can always extend to a continuous function $U \to Y$ for some open neighborhood of $A$? Or under what condition this is true? I ask this question because it seems that in the famous paper [Faisceaux Algébriques Cohérents] by Jean-Pierre Serre, he use this kind of conclusion to extend a section of a sheaf.
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1A well-known result of this kind is Tietze extension theorem. – hardmath Nov 27 '13 at 13:28
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I know the point now, since the projection of the sheaf to the base space is locally homeomorphism, thus this is a special problem. I think Tietze extension theorem cannot work here since there is no restriction on space here. I believe it's not true in general. – Lao-tzu Nov 27 '13 at 13:30
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A related Question at MathOverflow provides a sketch of a case where continuous extension is impossible. – hardmath Nov 27 '13 at 13:34
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For a toy counterexample, consider the three-point space $X = \{a,b,c\}$, where the open sets are $\emptyset$, $\{b\}$, $\{a,b\}$, $\{b,c\}$, and $X$. Then $\{a,c\}$ is closed, and can be mapped to the two-point discrete space $\{0,1\}$ by taking $a$ to $0$ and $c$ to $1$. But this map cannot be extended to $X$ (which is the only open neighborhood of $\{a,c\}$), since $X$ is connected.
This sort of trick can be used for many non-Hausdorff spaces mapping to two points.
As hardmath mentions, the statement applies to maps from normal spaces to the real numbers, by the Tietze extension theorem.
MartianInvader
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This and a more subtle (completely regular but not normal) example of Tietze extension theorem not holding were previously described here, Example where Tietze Extension fails?. – hardmath Nov 28 '13 at 02:21
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