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Let $\mathscr{F}$ be a sheaf on $X$, and $Y\subset X$ a subset. Define a presheaf $\mathscr{F}|_Y$ on $Y$ via the direct limit $$\mathscr{F}|_Y(V):=\lim_{V\subset U}\mathscr{F}(U),$$ where $V$ is an open subset of $Y$, and $U$ is an open subset of $X$. Clearly $\mathscr{F}|_Y$ is a sheaf if $Y$ is an open subset of $X$. But I can't think of an example where $\mathscr{F}|_Y$ is not a sheaf. Can anyone think of an example?

ashpool
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1 Answers1

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$\DeclareMathOperator{\sh}{Sh}$ Edit: The question originally asked about $F|_Y(V)=\varinjlim_{U\subset V}F(U)$, and my answer reflects this. It now looks like the OP is asking out the inverse image functor.

I claim that $F|_Y$ is the wrong thing to look at. In general, if $f:Y\to X$ is a continuous map between topological spaces (eg. the an embedding $Y\subset X$) then there is a natural functor $f_*:\sh(Y)\to\sh(X)$ given by $$ (f_* F)(U) = F(f^{-1}(U)) $$ In your case ($f:Y\to X$ is the embedding $Y\subset X$) one has $(f_* F)(V) = F(V\cap Y)$. Now, this functor isn't in the direction that you're looking for, but $f_*$ has a left adjoint $f^*:\sh(X)\to\sh(Y)$, which is characterized by the universal property $$ \hom_{\sh(Y)}(f^* F,G) = \hom_{\sh(X)}(F, f_* G) $$ for $F\in\sh(X)$, $G\in\sh(Y)$. (Note: in algebraic geometry one often writes $f^{-1}$ instead of $f^*$, but I'm just talking about sheaves of sets here.) Anyways, $f^* F$ is defined as follows: $$ (f^* F)(V) = \varinjlim_{U\supset f(V)} F(U) $$ So in your case ($Y\subset X$) one has $$ (f^* F)(V) = \varinjlim_{U\supset V} F(U) $$ which is very different than $\varinjlim_{U\subset V} F(U)$. For example, if $V$ has empty interior, than your definition yields $F|_Y(V) = *$ for all $V$.

Edit: Georges pointed out that I was overly hasty: my definition of $f^* F$ yields a presheaf, the sheafification of which is the "actual" $f^* F$. Anyways, here is an example of an inclusion $f:Y\to X$ and a sheaf $F$ on $X$ for which $f^{pre}F:V\mapsto\varinjlim_{U\supset f(V)} F(U)$ is not a sheaf.

Let $Y=\mathbb{R}$ with the discrete topology, $X=\mathbb{R}$ with the usual topology, and $f:Y\to X$ be the identity map. Then it is easy to see that if $F\in\sh(X)$ is the "sheaf of continuous functions," then $f^{pre} F$ is not a sheaf.

Daniel Miller
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  • The typo has been corrected, so the last part of your answer is outdated. The preceding parts seem valuable, though. – Daniel Fischer Jul 09 '13 at 17:41
  • I'm terribly sorry for the typo! – ashpool Jul 09 '13 at 17:51
  • @ashpool. No problem. The corrected definition gives a sheaf. – Daniel Miller Jul 09 '13 at 17:52
  • Dear Daniel, your definition of $f^* F$ is not correct: the rule $V\mapsto f^{pre} F=\varinjlim_{U\supset f(V)} F(U)$ in general only defines a presheaf and not a sheaf: can you find an example confirming my assertion? The inverse image sheaf $f^* F$ is the sheaf associated to that presheaf $f^{pre} F$. – Georges Elencwajg Jul 09 '13 at 21:38
  • Dear Georges: I guess that's what happens when all you ever need in proofs is the underlying presheaf. I've included an example showing that $f^{pre} F$ isn't always a sheaf in my answer. – Daniel Miller Jul 09 '13 at 22:37
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    Dear Daniel, your reaction to my comment is quite praiseworthy: you corrected your answer and invented a quite ingenious example: +1. – Georges Elencwajg Jul 10 '13 at 06:14
  • I wonder if there is an example where the inclusion $Y\rightarrow X$ is an embedding, so that $Y\subset X$ is actually a subspace. – ashpool Jul 13 '13 at 15:19
  • @Daniel: the example you gave is a continuous map $Y\rightarrow X$ which is not an embedding. So one cannot regard $Y$ as a subspace of $X$. – ashpool Jul 13 '13 at 16:06