$i^{-1}\mathcal F$ presheaf is a sheaf?

Question

Let $X$ be a closed subset of $Y$ and $i: X\to Y$ be the embedding map. Is it true that for a sheaf $\mathcal F$ its $i^{-1}\mathcal F$ presheaf is a sheaf?

Answer 1

Let $U_j$ be an open covering of $U$, an open of $X$. A section of $i^{-1}F$ over $U_j$ is the data of an open $V_j$ of $Y$ such that $V_j \cap X = U_j $ and a section $s_j \in F(V_j) $, identified over restriction to "smaller and smaller $V_j$'s ".

Now take compatible sections over each $U_j$. This means having pairs as above $(V_j, s_j) $ such that for any $i, j$ there exist a $U_i \cap U_j \subset V_{ij} \subset V_i \cap V_j$ with the property that

$$ s_i | V_{ij} = s_j | V_{ij} $$

This is the problem: we have no obvious candidate to take here. Here:

https://mathoverflow.net/questions/75580/does-one-need-to-sheafify-when-defining-the-inverse-image-of-a-sheaf-with-respec

There is a comment with a counterexample, in which it is actually not a sheaf. However, I think that if you have a closed embedding of manifolds $i:N \to M$, then it should be true. I can think about this if you are interested in.