Example of a restriction of sheaf not being a sheaf 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?
 A: $\DeclareMathOperator{\sh}{Sh}$
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. 
