2

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.

Bryan261
  • 854

1 Answers1

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).

  • 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
  • this is because U is maximal (in particular cofinal) wrt the partial order. I have extended some explanations, is it clear now? –  Jul 13 '14 at 18:08
  • 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
  • this the standard property of direct limits mentioned above. are you not familiar with direct limits? it is proven e.g. here: http://math.stackexchange.com/questions/334523/bijection-between-direct-limits –  Jul 13 '14 at 18:16
  • Yes, i'm not enough familiar with direct limits. Sorry. – Bryan261 Jul 13 '14 at 18:18