Is the inverse image of a sheaf along a embedding always a sheaf?

Question

Let $i: X\to Y$ be a embedding map between topological spaces, given any sheaf $\mathcal F$ on $Y$, is $i^{-1}\mathcal F^{\text{pre}}$ automatically a sheaf?

Here the presheaf $i^{-1}\mathcal F^{\text{pre}}$ is defined by $i^{-1}\mathcal F^{\text{pre}}(U):=\varinjlim_{V\supset i(U)}\mathcal F(V)$.

Answer 1

For a simple counterexample, let $Y=\{a,b,c\}$, with the topology generated by the sets $\{a,b\}$ and $\{b,c\}$. Let $X=\{a,c\}$, and let $\mathcal{F}$ be the constant sheaf $\mathbb{Z}$ on $Y$. Then $i^{-1}\mathcal{F}^{pre}(X)=\mathbb{Z}$, but $X$ is discrete so when you sheafify you get $i^{-1}\mathcal{F}=\mathbb{Z}\oplus\mathbb{Z}$.

For a Hausdorff counterexample, you can let $L=[0,\omega_1]$ be the closed long line, $Y=[0,1]\times L\setminus\{(1,\omega_1)\}$ (i.e., a connected version of the deleted Tychonoff plank), and $X=\{1\}\times [0,\omega_1)\cup[0,1)\times\{\omega_1\}$. Again taking $\mathcal{F}$ to be a constant sheaf, $i^{-1}\mathcal{F}^{pre}$ takes the wrong value on $X$ because $X$ is disconnected, but any open neighborhood of $X$ in $Y$ contains a connected neighborhood of $X$.

Answer 2

In general, it is not true. It is certainly true in the following special cases:

• $X \subset Y$ is open.
• $X$ is a point.

A counterexample in the general case is quite easy: Take $Y$ to be any irreducible space, $X$ to be two isolated points and $\mathcal F$ a constant presheaf on $Y$.

Since $Y$ is irreducible, the constant presheaf is a sheaf on $Y$. But it is not a sheaf von $X$.