Let $X$ be a topological space containing two closed points $x,y$ and let $i : \{x,y\} \to X$ denote the inclusion map. Notice that $\{x,y\}$ carries the discrete topology. Let $F$ be a sheaf on $X$. Then $i^{-1} F$ is a presheaf on $\{x,y\}$ which is given by $(i^{-1} F)(\emptyset)=1$ (the terminal set), $(i^{-1} F)(\{x\}) = F_x$ (the stalk at $x$), $(i^{-1} F)(\{y\})=F_y$ (the stalk at $y$) and $$(i^{-1} F)(\{x,y\}) = \varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U).$$ Why is, $i^{-1} F$ a sheaf if and only if the canonical map $$\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y.$$ is an isomorphism ?.

Thanks a lot to all of you.

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    $\begingroup$ The only nontrivial covering on the discrete set $\{x,y\} $ is $\{x,y\}=\{x\} \cup \{y\} $. So for a presheaf $G$ on $\{x,y\} $ to be a sheaf, the only thing to verify (apart from $\#G(\varnothing)=1$) is $G(\{x,y\} )\stackrel{\sim}{\rightarrow}G(\{x\} )\times G(\{y\} )$. This is the condition you want for $i^{-1}F$. $\endgroup$ – abx Jul 16 '14 at 15:45
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    $\begingroup$ Posting the same question at the same time on two sites is equivalent to calling two pizza places and ordering pizza, just to see which one gets to your place first. $\endgroup$ – Asaf Karagila Jul 16 '14 at 19:46
  • $\begingroup$ Looks like being copied from here ;) $\endgroup$ – principal-ideal-domain Jul 26 '14 at 12:50

A presheaf is a sheaf iff it is mono and conjunctive (maybe there is other terminology) For mono: Say that $f \in i^{-1}F(X)$ and if $r_x(f) \in i^{-1}F(\{x\})$ and $r_y(f) \in i^{-1}F(\{y\})$ are both zero then $f$ must be zero. this means that the map given by restrictions $$i^{-1}F(X) \rightarrow i^{-1}F(\{x\})\times i^{-1}F(\{y\})$$ is injective.

For conjuctive we get that the same map is surjective for if $f_x \in i^{-1}F(\{x\})$ and $f_y \in i^{-1}F(\{y\})$ then they agree on the common domain $\{x\} \cap \{y\}$ trivially so and there must be $f\in i^{-1}F(X)$ which resticts to $f_x$ and $f_y$.

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  • $\begingroup$ Thank you for this answer. Could you explain to me please, what does mean that : $ f \in \varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) $ in general ?. Thank you. $\endgroup$ – Bryan261 Jul 16 '14 at 15:36
  • $\begingroup$ I am not sure what you mean, its a direct limit ? It just means that $f\in F(U)$ for some $U$ ? $\endgroup$ – Rene Schipperus Jul 16 '14 at 15:40
  • $\begingroup$ But $ f $ is an equivalence class not only an element of $ F(U) $ as i know. I only want to recall the definition of an element of $ \varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) $. Thanks a lot. $\endgroup$ – Bryan261 Jul 16 '14 at 15:44
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    $\begingroup$ You should think of a direct limit $\lim_{U} F(U)$ as being the all $f$ such that for some $U$, $f \in F(U)$, naturally once $f \in F(U)$ we then identify $f$ with its image in $F(V)$ for $V \subseteq U$. Note however that $f, g\in F(U)$ may become equal restricted to a smaller set. Thats how I think about it. $\endgroup$ – Rene Schipperus Jul 16 '14 at 15:49
  • $\begingroup$ Ok, thank you very much. :-) $\endgroup$ – Bryan261 Jul 16 '14 at 15:51

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