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.

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    $\begingroup$ If $j$ is an open immersion things simplify a lot. $\endgroup$ – Zhen Lin Jul 13 '14 at 17:03
  • $\begingroup$ Why is : $ j^{-1} \mathcal{F} = \mathcal{F}_{|U} $, in this case ? $\endgroup$ – Bryan261 Jul 13 '14 at 17:13
  • $\begingroup$ 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 $f^{-1}\mathscr{F}$. $\endgroup$ – 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).

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  • $\begingroup$ I don't understand why : $\mathscr{F}|_X(U)=\varinjlim_{U \subset V} \mathcal{F} (V) =\mathcal{F}(U)$. $\endgroup$ – Bryan261 Jul 13 '14 at 17:51
  • $\begingroup$ this is because U is maximal (in particular cofinal) wrt the partial order. I have extended some explanations, is it clear now? $\endgroup$ – user164074 Jul 13 '14 at 18:08
  • $\begingroup$ 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 $. $\endgroup$ – Bryan261 Jul 13 '14 at 18:11
  • $\begingroup$ this the standard property of direct limits mentioned above. are you not familiar with direct limits? it is proven e.g. here: math.stackexchange.com/questions/334523/… $\endgroup$ – user164074 Jul 13 '14 at 18:16
  • $\begingroup$ Yes, i'm not enough familiar with direct limits. Sorry. $\endgroup$ – Bryan261 Jul 13 '14 at 18:18

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