3
$\begingroup$

I came across two different definitions of sheafification and I'm not sure how they are equivalent. One of them is here: About the sheafification

Another one is from Tennison's sheaf theory: Given a presheaf $F$ over $X,$ we construct the sheaf space $LF:=\sqcup_{x\in X}\mathscr{F_{x}}$ which is a disjoint union of stalks and the continuous map is the natural projection $p:LF\rightarrow X $. The open sets are $s[U]=${ $s_{x}\in LF: x\in U$}. Then the sheafification is $\Gamma LF,$ where we construct the sheaf of sections on the sheaf space $LF.$ $\Gamma LF(U)=${$\text{continuous maps} \ \ \sigma :U\rightarrow LF \ \ \text{s.t.} \ \ p(\sigma)=id_{U} $}. So for example, the map $\hat{s}:U\rightarrow LF(U)$ where $x \mapsto s_{x}\in \mathscr{F_{x}}$ would be in $\Gamma LF(U)$. Can someone explain how they are equivalent?

$\endgroup$
  • $\begingroup$ They are exactly the same. You just have to keep in mind there is only one left adjoint to the forgetful functor from sheaves to presheaves. This means that the sheaffification is the best approximation in the sense that any other arrow going to a sheaf will factorize uniquely through the sheaffification. $\endgroup$ – user40276 Aug 6 '14 at 23:39
2
$\begingroup$

Here's a concrete way to go back and forth between the descriptions.

A continuous function $\sigma:U \rightarrow LF$ satisfying $p\circ \sigma = \mathbb 1$ corresponds to the element $(\sigma (x)) \in \prod_{x\in U} F_x$. To show that $(\sigma(x)) \in F^{\#}(U)$, let $a \in U$ and let $U'\subset LF$ be so that $\sigma(a) \in U'$. Since basic open sets of $LF$ are of the form $s[V]$ for $s \in F(V)$, we may assume $U' = s[V]$. By continuity of $\sigma$, there exists a neighborhood $W$ of $a$ so that $\sigma(W) \subset s[V]$. But this implies $\sigma(b)=s_b$ for all $b \in W$. Since this holds for all $a \in U$, $(\sigma(a))$ satisfies $(*)$.

For the other direction, take $(s_x)\in F^{\#}(U)$ and let $\sigma$ be the section of $p$ over $U$ defined by $x \mapsto s_x$. We must verify $\sigma$ is continuous. For any $x \in U$ and any basic open set $t[V]\subset LF$ with $s_x \in t[V]$, $(*)$ implies that there exists $W \subset V$ so that $s_y = t_y$ for all $y \in W$. This shows that $x\in\sigma(W) \subset t[V]$, and hence $\sigma$ is continuous.

To completely nail down the equivalence of the definitions, you'll want to check that these identifications of $\Gamma LF(U)$ and $F^{\#}(U)$ for $U \subset X$ open establish an isomorphism of sheaves (i.e. that the maps $\Gamma LF(U) \rightarrow F^{\#}(U)$ are morphisms and the appropriate diagrams involving restrictions commute).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.