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I have started to read an english translation of Serre's FAC. Immediately a sheaf is defined. The more categorical definition given on wikipedia actually makes more sense to me, but I would like to understand this one as it is written since I will be working through this paper.

I give the definition, then pose some questions.

Definition:

Let $X$ be a topological space. A sheaf of abelian groups on $X$ (or simply a sheaf ) consists of:

(a) A function $x \to \mathscr{F}_x$, giving for all $x \in X$ an abelian group $\mathscr{F}_x$,

(b) A topology on the set $\mathscr{F}$, the sum of the sets $\mathscr{F}_x$.

If $f$ is an element of $\mathscr{F}_x$, we put $\pi(f) = x$; we call the mapping of $\pi$ the projection of $\mathscr{F}$ onto $X$; the family in $\mathscr{F} \times \mathscr{F}$ consisting of pairs $(f,g)$ such that $\pi(f) = \pi(g)$ is denoted by $\mathscr{F}+\mathscr{F}$.

Having stated the above definitions, we impose two axioms on the data (a) and (b):

(I) For all $f \in \mathscr{F}$ there exist open neighborhoods $V$ of $f$ and $U$ of $\pi(f)$ such that the restriction of $\pi$ to $V$ is a homeomorphism of $V$ and $U$.(In other words, is a local homeomorphism).

(II) The mapping $f \mapsto -f$ is a continuous mapping from $\mathscr{F}$ to $\mathscr{F}$, and the mapping $(f, g) \mapsto f + g$ is a continuous mapping from $\mathscr{F}+\mathscr{F}$ to $\mathscr{F}$.

Questions:

1) At the very beginning of this definition, what are we taking $\mathscr{F}$ to be? It seems we are referring to it indirectly as the category of abelian groups. In this definition is $\mathscr{F}$ just some unspecified class of abelian groups? They refer to $\mathscr{F}$ as a set in (b), but I don't believe the category of abelian groups is small. What should I be taking $\mathscr{F}$ as?

2) After this definition, which thing(s) exactly is (are) the sheaf? Is it the pair $(f,\tau)$ where $f$ is the function from (a) and $\tau$ the topology from (b)? The function from (a) seems to be playing the role of the functor from the definition on wikipedia, except that the functor pairs open sets of $X$ to objects, not points of the space $X$.

3) In part (b), a priori of the rest of the definition, is it just stating that any topology can be on $\mathscr{F}$? Why do they say 'the sum of the sets $\mathscr{F}_x$'.

4) In I and II, how am I to make sense of $-f$ and $f+g$ if $f,g$ are abelian groups? does this have to do with the topology we put on the collection of sets?

5) Is this definition actually equivalent to the one on wikipedia? Of course this is just for abelian groups, and the one on wikipedia allows the target category to be Sets, Rings, etc.. But in the case of the target category being abelian groups, are they equivalent?

EDIT:

  • Perhaps question 3 actually answers question 1. Are they defining the 'set' $\mathscr{F}$ to be the sum of all the images of the function? In what set theoretic way are we summing them? The disjoint union seems to be plausible?

  • Also, -f and f+g do make sense, I just realized that $f,g$ are elements of the group $\mathscr{F}_x$.

  • In light of these edits, can I take $$\mathscr{F} = \bigcup_{x \in X} (\{x\} \times \mathscr{F}_x) \hspace{2mm} ?$$ then the mapping $\pi$ mentioned in the definition would really be mapping from $(x,f) \mapsto x$? Although maybe trivial at this level, this seems important since in the case of the constant sheaf the same group element would be going to each $x \in X$ but in this disjoint union that would still be well defined.

  • The definition of the map $\pi$ seems very 'not universal' since it depends on the elements of the groups?

  • The definition of $\mathscr{F} + \mathscr{F}$ seems very strange, it is just forming all tuples of elements that came from the same group?

Prince M
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    For 'sum' in point (b), take disjoint union. Compare the corresponding definition of, say, a vector bundle over a nice manifold. – anomaly Apr 29 '18 at 23:57

1 Answers1

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1) In (b), "sum" means disjoint union. So $\mathscr{F}$ is the union of the groups $\mathscr{F}_x$, which are the "stalks" of the sheaf.

2) In this definition, the sheaf is the space $\mathscr{F}$, with the appropriate topology. It is also common to say that the sheaf "is" the functor sending an open subset $U \subset X$ to the set $\mathscr{F}(U)$ of continuous sections $U \to \pi^{-1}(U)$, which in fact has the structure of an abelian group by axiom (II).

3) Any topology on $\mathscr{F} = \bigsqcup_x \mathscr{F}_x$ satisfying axioms (I) and (II) is a valid topology defining a sheaf.

4) You seem to have figured this one out already.

5) Yes. See 2) for what the functor should be.

You should keep in mind the simplest example of a sheaf, given directly below the definition, which is $\mathscr{F} = X \times G$ for an abelian group $G$, where $G$ is given the discrete topology. So here $\mathscr{F}_x = G$ for every $x$. Translating to the language of a functor on open sets, this is the sheaf of locally constant $G$-valued functions.

nkm
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  • Thanks! I just worked through verifying the constant sheaf satisfies the definitions. I am hung up on verifying that (f,g) -- > f+g is continuous, what is the topology we are taking on $\mathscr{F}+\mathscr{F}$? is it a subspace topology of the product topology? – Prince M Apr 30 '18 at 01:17
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    Yes, it should have the subspace of the product topology. So a basis is open sets of the form $U \times {(f,g)}$ for open $U \subset X$, $f,g \in G$ – nkm Apr 30 '18 at 01:21
  • So starting with my definition and then verifying the functorial definition, is $\pi^{-1}(U)$ really how the functor that takes $U$ to its set of sections would be implemented? It seems that starting with this definition we would want to define the functor to take $U$ to the set of all continuous map $s: U \to \mathscr{F}$ such that $\pi \circ s = id_U$, which would be, I think, a subset of the powerset of $\pi^{-1}(U)$? And as mentioned that set of maps $s$ would have an abelian group structure. – Prince M May 01 '18 at 00:25
  • And the two definitions are the same, not in the sense that they define the same 'object', since in one, the actual $\textit{sheaf}$ is the disjoint union / collection of stalks, and in the other the $\textit{sheaf}$ is the functor that pairs open sets to its sections, but they are the same in the sense that they specify and assign the same data to the topological space? One could still prove they are the same by showing that if you start with one you can naturally meet the specifications of the other? Perhaps my definition above fits the anology to a real life 'sheaf' better? – Prince M May 01 '18 at 00:44
  • Yes, the functor takes $U$ to the set of all continuous sections $s: U \to \mathscr{F}$ ("section" meaning $\pi \circ s = id_U$. The arrow $U \to \pi^{-1}(U)$ in my answer was supposed to be such a section $s$, not an indication that the functor sends $U$ to the set $\pi^{-1}(U)$. – nkm May 01 '18 at 02:11
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    The second question is a bit more philosophical, but I guess I would say they are formally different definitions, but starting from either one you can naturally create the other one. Kind of like how some people define "functions" $f: X \to Y$ as subsets $G_f \subset X \times Y$ such that $S_f \cap ({x} \times Y)$ contains exactly one point for each $x \in X$. More common is to call $G_f$ the graph of the function $f$, but you can naturally pass from one to the other, so they really consist of the same data. – nkm May 01 '18 at 02:14
  • Oh ok, that makes sense, you were just saying U maps into $pi^{-1}(U)$. – Prince M May 01 '18 at 02:14
  • Thank you for all your help. Look for more advanced sheaf questions from me in the near future! – Prince M May 01 '18 at 02:20
  • In Serres paper he says that if U is open in X, then s(U) is open in the sheaf, why is this? – Prince M May 02 '18 at 03:55
  • I see it should have to do with axiom I but I don’t quite see – Prince M May 02 '18 at 04:05
  • Let $x \in U$, and $V \ni s(x)$ a neighborhood given by axiom I. By continuity, $W_x = s^{-1}(V)$ is an open neighborhood of $x$ in $U$. The axiom I property tells you that $s$ maps $W_x$ homeomorphically to an open subset of $V$, which is then open in $\mathfrak {F} $. Openness is a local condition, so $s $ is an open map. – nkm May 02 '18 at 04:52
  • Ok, I see now that $s$ carries $W_x$ homeomorphically to some open subset of $V$. Since $s$ is a homeomorphism restricted to $W_x$ it must be an open map on the restriction, this is enough to say $s$ is an open map? – Prince M May 02 '18 at 06:32
  • Yeah that's enough. Openness is local. $s(U) = \bigcup_{x\in U} s (W_x) $ is open. – nkm May 02 '18 at 07:08
  • Of course, very good! – Prince M May 02 '18 at 07:13
  • Just make sure, to verify s carries W_x honeomorphically to an open subset of V, I want to check that when I restrict the homeomorphism that comes for free with V, restricted to the subset of V, it’s inverse is s? – Prince M May 03 '18 at 20:37
  • Is it actually W_x that s maps homeomorphically or would it be W_x intersect pi(V)? – Prince M May 03 '18 at 20:57
  • If $y \in W_x = s^{-1}(V)$, then $y = \pi(s(y)) \in \pi(V)$, so $W_x \subset \pi(V)$. – nkm May 03 '18 at 21:37
  • That clears everything up. That was one of the things I thought of, but I got lost in the symbols! – Prince M May 03 '18 at 21:44