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

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 $i^{-1}\mathcal F^{\text{pre}}(U):=\varinjlim_{V\supset i(U)}\mathcal F(V)$.

• Did you mean the direct rather than inverse limit? – mad_algebraist Feb 18 '16 at 2:21
• @mad_algebraist Yes, I mean direct limit, thak you for pointing out that. – Censi LI Feb 18 '16 at 6:27

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 $X$.

• Than you, nice example. Do you think it holds when $Y$ is Hausdorff? – Censi LI Feb 18 '16 at 6:59
• I've added a Hausdorff counterexample. Probably there is a less complicated example as well, but that's what I could come up with off the top of my head. – Eric Wofsey Feb 18 '16 at 7:05

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 $X$.

• By "isolated points", you mean closed points? Or more generally, a pair of points neither of which is a specialization of the other. – Eric Wofsey Feb 18 '16 at 6:42
• Yes. I just want to make sure that in the subset topology of $X$, both points are clopen. – MooS Feb 18 '16 at 6:44
• Than you, very insightful. Do you think it holds when $Y$ is Hausdorff? – Censi LI Feb 18 '16 at 6:59