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Let $F$ be a presheaf on a topological space $X$ of some category of "sets with structure." In Borel's Linear Algebraic Groups, he gives the following explanation for how to construct the associated sheaf $F'$:

Roughly speaking, $F'$ can be constructed in two steps. First, define $F_1(U)$ to be $F(U)$ modulo the equivalence relation which relates $s$ and $t$ if their restrictions agree on some open cover of $U$. Then form $F'(U)$ by "adding" to $F_1(U)$ all elements obtainable from compatible local data on some open covering of $U$. This process makes sense thanks to step 1.

Unfortunately, this process does not make sense. I permit no thanks to step 1. Anyway, this differs from other constructions of sheafification which I have seen before. The main one I am familiar with is to define $F'(U)$ to be the set of functions $f$ from $U$ into the disjoint union of stalks $F_x : x \in U$ such that the following holds: for each $x \in U$, $f(x) \in F_x$, and there is an open covering $U_i$ of $U$, and sections $s_i \in F(U_i)$, such that $f(x) = (s_i,U_i)_x$ for each $x \in U_i$.

Is there a nice way to understand what Borel is talking about? I just don't get it. Is there a nice way to relate these two notions of sheafification together? (besides universal property shenanigans)

D_S
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    What part doesn't make sense? – Hoot Jan 23 '16 at 21:14
  • Step 1 does make sense. On step 2, you're supposed to run through all possible open covers $U_i$ of $U$ (?) and for every such cover, and every $s_i \in F(U_i)$ such that $s_{i|U_i \cap U_j} =s_{j|U_i \cap U_j}$ for all $i, j$, you add to $F_1(U)$ an element $s$ subject to..the same equivalence relations? Then you can somehow make that into an abelian group/ring/module. I don't even know if that's right. I can't give you a good answer for why it doesn't make sense, it just doesn't lol – D_S Jan 23 '16 at 21:20
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    A presheaf is a sheaf if and only if $$F(U) \to \prod_i F(U_i) \rightrightarrows \prod_{i,j} F(U_i \cap U_j)$$ is an equalizer diagram. This can fail in two ways: (1) the first map need not be injective; (2) two things that are the same in $\prod_{i,j} F(U_i \cap U_j)$ do not have to be in the image of $F(U)$. To get a sheaf, Borel corrects (1) in step one, then corrects (2) in step two. See Ch. II, §1 in Artin's Grothendieck Topologies, or Ch. III, §5 in Mac Lane/Moerdijk's Sheaves in Geometry and Logic. – Takumi Murayama Jan 24 '16 at 00:31
  • @ElíasGuisadoVillalgordo Do you have a good reason why one has to do two equivalence relations? It somehow seems like we have to say "identify two things when they agree on a cover" twice. Put differently: what fails in the second step if the presheaf is not separated? – red_trumpet Oct 23 '23 at 20:59
  • @red_trumpet I moved the discussion to an answer below because it was starting to get lengthy from my side . – Elías Guisado Villalgordo Oct 24 '23 at 08:42

2 Answers2

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For a covering $\mathcal{V}$ of $U$, define $A_\mathcal{V}(U)\subset\underset{V\in\mathcal{V}}\prod F_1(V)$ as the subgroup of elements that agree on the intersection of any two $V,W\in\mathcal{V}$.

The set $\Phi$ of covers of $U$ is ordered by refinement, and if $\mathcal{V}<\mathcal{V'}$ then you have many inclusion maps $i_{\mathcal{V},\mathcal{V'}}:\mathcal{V}\to\mathcal{V'}$, which turn it into a filtered category.

To every homset $\operatorname{Hom}(\mathcal{V},\mathcal{V'})$, you can associate one restriction map $\rho_{\mathcal{V'},\mathcal{V}}:A_\mathcal{V'}\to A_\mathcal{V}$, so you can finally define the sheafification of $F$ as $F'=\underset{\mathcal{V}\in\Phi}{\operatorname{colim}} A_\mathcal{V}$.

Note that the only subtle point is how you define the restriction maps $\rho$, since each one is really induced by the all homset, not just a single inclusion map $i$.

You can find this construction in Voisin's Hodge Theory and Complex Algebraic Geometry, I lemma 4.4

Dinisaur
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Even though the OP explicitly asked for an non-formal answer, here's the formal answer (in case it's useful to anyone): Given a separated sheaf $G$, define $G'$ to be the separated presheaf where a section over $U$ is a collection of pairs $(U_i,s_i)$, where $U=\bigcup_iU_i$ is an open cover and $s_i\in G(U_i)$ agree on intersections, modulo the equivalence relation $(U_i,s_i)\sim(V_j,t_j)$ if $s_i|_{U_i\cap V_j}=t_j|_{U_i\cap V_j}$ for all $i,j$. Then: (1) check the map $F\to F_1$ is universal among morphisms from $F$ into separated presheaves, (2) check the map $G\to G'$ is universal among morphisms from $G$ into sheaves (here ). Using (1)+(2), one verifies that the composite $F\to F_1\to F'=(F_1)'$ is universal among morphisms from $F$ into sheaves.

red_trumpet asked on a comment:

Do you have a good reason why one has to do two equivalence relations? It somehow seems like we have to say "identify two things when they agree on a cover" twice. Put differently: what fails in the second step if the presheaf is not separated?

What exactly fails is transitivity of the relation $(U_i,f_i)\sim(V_j,g_j)$, for if we also have $(V_j,g_j)\sim(W_k,h_k)$, then $f_i|_{U_i\cap V_j\cap W_j}=g_j|_{U_i\cap V_j\cap W_j}=h_k|_{U_i\cap V_j\cap W_j}$ for all $i,j,k$, and this implies $f_i|_{U_i\cap W_k}=h_k|_{U_i\cap W_k}$ for all $i,k$ only when the presheaf is separated. The need to introduce step 1 is because there are non-zero presheaves whose sheafifications are trivial (such presheaves are necessarily non-separated); whereas step 2 gives a map $G\to G'$ that is always injective (in particular, in abelian separated presheaves, $G\neq 0$ implies $G'\neq 0$). Let's call a practically zero section to a section of an abelian presheaf whose restrictions to the sets of some open cover vanish. More descriptive slogans are, for step 1, "mod out by practically zero sections"; for step 2, "force the gluing of sections that agree on intersections."

I guess the reason why this construction of the sheafification is not that extended on the literature is because it makes one work with two equivalence relations at the same time. I think the instance where this presentation is most useful is when sheafifying separated presheaves (e.g., the quotient presheaf of an abelian sheaf by an abelian subsheaf is always separated), so one deals only with one equivalence relation. It's nice to think about Cartier divisors this way; this is what Görtz, Wedhorn, Algebraic Geometry I, 2nd ed. does in Definitions 11.20 and 11.26 (1).