# The inverse image of a sheaf

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.

• If $j$ is an open immersion things simplify a lot. – Zhen Lin Jul 13 '14 at 17:03
• Why is : $j^{-1} \mathcal{F} = \mathcal{F}_{|U}$, in this case ? – Bryan261 Jul 13 '14 at 17:13
• I think $f^{-1}\mathscr{F}$ is already the sheafification of that what you defined to be $f^{-1}\mathscr{F}$. In my textbook your direct limit is defined to be $f^{+}\mathscr{F}$ and its sheafification $f^{-1}\mathscr{F}$. – principal-ideal-domain Jul 26 '14 at 16:29

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).
• I don't understand why : $\mathscr{F}|_X(U)=\varinjlim_{U \subset V} \mathcal{F} (V) =\mathcal{F}(U)$. – Bryan261 Jul 13 '14 at 17:51
• Not yet. Sorry. I want to prove clearly why : $\mathcal{F} ( U ) = \displaystyle \lim_{ \longrightarrow U \subset V } \mathcal{F} ( V )$. I d'ont know the meaning of cofinal subset of $I$. – Bryan261 Jul 13 '14 at 18:11