Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question is an attempt to understand the proof of lemma 6.17.3 (page 164) in the Stacks Project: http://www.math.columbia.edu/algebraic_geometry/stacks-git/book.pdf

This lemma states:

Let $F$ be a presheaf of sets on $X$. Any map $F\to G$ into a sheaf of sets factors uniquely as $F \to F^{S} \to G$ (when $F^{S}$ denotes the sheafification of $F$).

The part that I'd like to understand is: why is the above map $F^{S} \to G$ unique?

share|improve this question
A sheaf map is determined by the map on each stalks, and the stalk maps induced by $F \to G$ are the same as those of $F^S \to G$. –  Soarer Jul 25 '11 at 17:46
add comment

3 Answers

up vote 1 down vote accepted

You could look at any proof of the existence of the associated sheaf for sheaves of sets, abelian groups... For instance: for sheaves of abelian groups, see B. Iversen, Cohomology of sheaves, Springer Universitext.

The key point is that another morphism of sheaves $F^S \longrightarrow G$ such that composed with the universal map $F\longrightarrow F^S$ gives you your original $F \longrightarrow G$, would induce the same morphism on all stalks $F_x = F^S_x \longrightarrow G_x$ for every point $x\in X$. Then, as morphisms of sheaves are determined by their induced maps on stalks (Iversen, lemma 2.2 (i)), they would agree.

share|improve this answer
add comment

Another perspective is that the sheafification functor $\mathbf{a}$ is left adjoint to the inclusion functor $\mathbf{i}$ from sheaves to presheaves.

We therefore get a one-to-one correspondence between maps of presheaves $F \to \mathbf{i}G$ and maps of sheaves $\mathbf{a}F \to G$.

The factorization you mention is $F \to \mathbf{ia} F \to \mathbf{i}G$; the first arrow is the unit of the adjuction, and the second arrow applies $\mathbf{i}$ to the latter arrow above. Because the inclusion is full and faithful, we still have the one-to-one correspondence.

share|improve this answer
add comment

Choose an open set $U \subset X$, and write $f:F\rightarrow G$. If we have some element $a\in F(U)$, then its image $f_U(a)\in G(U)$ is determined by its stalks since $G$ is a sheaf. So $f$ is determined by its action on stalks. Since $F^S$ is a sheaf, we then get a unique map $F^S \rightarrow G$ determined by that action.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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