The inverse image of a sheaf

Question

By definition, the inverse image of the sheaf $ \mathcal{F} : \mathrm{Ouv} (Y) \to \mathrm {Set} $ is the sheaf associated to the presheaf $ f^{-1} \mathcal{F} : \mathrm{Ouv} (X) \to \mathrm{Set} $ defined by $ f^{-1} \mathcal{F} (U) = \displaystyle \varinjlim_ {f (U) \subset V} \mathcal{F} (V) $. How does $ f^{-1} \mathcal{F} $ become, when $f: X \to Y $ is an inclusion map ? Thanks a lot.

Answer 1

We have $f^{-1}\mathscr{F}=\mathscr{F}|_X$ if $X$ is open in $Y$. What you are looking for is that you can compute the direct limit over cofinal subsets, meaning that $\varinjlim_{i\in I} A_i.=\varinjlim_{j\in J} A_j$ if $J$ is a cofinal subset of $I$. For this reason we have for $U$ open in $X$ that $f^{-1}\mathscr{F}(U)=\varinjlim_{U \subset V} \mathscr{F} (V) =\mathscr{F}(U)=\mathscr{F}|_X(U)$, since $U$ (also open in $Y$ since $X$ is open) is maximal in the partial order over which you are taking the direct limit (the open nbhds of $U$ in $Y$, partially ordered by REVERSE inclusion).