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Motivation between sheafs and category theory

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Jose Capco

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Dec 8, 2005, 6:09:49 AM12/8/05
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Dear NG,

Ok, I know that there is a clear relation between sheaf theory and
category theory and I had like to make myself comfortable in the way
sheaf theory is done using category-theoretical definitions and notions
(I learned sheaf theory the "classical" way, in classical algebraic
geometry.. without any inital knowledge on category theory and etc.)

I am for instance aware that one can define sheaves using equalizers in
category
i.e. one can define a sheaf to be a functor

F: X^op -> Groups/Sets/R-modules/whatnot

(The functor above just say that the sheaf must first be a presheaf!)
where X^op is a category corresponding to the topol. space X with
containment as morphism (i.e. if U and V are open with
V\subset U then we define a morphism U->V .. this is just the dual of X
with inclusion as morphisms)

such that there is an equalizer (this is just the "glued functions")
for any covering {U_i} of U

FU ---> \prod F(U_i)

for the parallel map

\prod_{i} FU_i ===> \prod_{i,j} F(U_i\cap U_j)

where the first map say p: \prod FU_i ---> \prod F(U_i\cap U_j)
takes an f_i to f_i|(U_i\cap U_j) and the second takes f_j to
f_j|(U_i\cap U_j)

The condition above ensures the two conditions of sheafs that one
learns in classical algebraic geometry (i.e. the identity condition.. I
think this comes from the universal property of the equalizer, and the
glueing condition)


There is probably a better way to define a sheaf using category
theory.. but well this is what I have now.

But now that I know this, how does it help my understand of sheaf
theory. What things will get easier to perform with these extra
category theory tools?

Sincerely,
Jose Capco

PS: I got that definition from a book.. but I am beginning to doubt
something here..I think for the definition to match properly with what
I have learned in classical alg. geom. the functor should be
surjective (I forgot already if it should be called "full" or
"faithful")

Lukas Horosiewicz

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Dec 8, 2005, 6:40:13 AM12/8/05
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While I haven't done almost any cohomology I heard that this makes life
easier there. Also, why should the functor be full?

Lukas Horosiewicz

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Dec 8, 2005, 6:52:27 AM12/8/05
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The thing is (I'm going to write what you wrote again, basically), a
presheaf is a contravariant functor from X (objects being the open sets
of X and morphisms the inclusion maps) to a nice enough category say
Grp/Ring etc (I think one also wants that the functor applied on the
empty set of X should be an initial object but this follows from the
equalizer diagram applied to the empty covering does it not? Otherwise
I think we get some minor problems with the constant sheafs). Now you
call that functor a sheaf if it further is the limit of your diagram,
ie if it is the equalizer (another way of saying that it's a sheaf is
that it sends colimits to limits [note that X as a category doesnt
necessarily need to admit (co)products but we can embedd it in such a
category]).

This agrees with what you said, but the full thingy you require seems
to be weird, would we not want the constant sheafs to be sheafs?

Jose Capco

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Dec 8, 2005, 7:15:43 AM12/8/05
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not full sorry.. I meant that it must be full on a subcategory of the
groups/sets/rings.. etc. .. If you know what I mean.. I thought because
we use the equalizer.. then I do see the the universal property of the
equalizer yields the "identity property", but there are extra property
involved too ...like if there is ANY group G (I deal with sheafs of
groups here) such that

and a morphism f: G ---> \prod F(U_i) such that

pf=qf then there is a unique mapping
g: G --> F(U)
(by the universal property of equalizers) such that eg=f

If G where F(U) this would only mean the identity theorem.. but if G
where ANY group.. then I dont see this in the "classical" definition of
sheafs (or at least I dont see how this can be derived from the
"classical" definition).. and I get even more confused if G were say
F(V) for V a subset of U.. is this even possible? ... and .. what do
you mean by "colimits are sent to limits"?

Lukas Horosiewicz

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Dec 8, 2005, 7:28:36 AM12/8/05
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I do not have the time to think through what you mean because I have a
lecture to attend in 5 minutes. However I think I can explain what I
mean with colimits to limits. Basically gluing sets can be thought of
as a colimit since the presheaf functor is contravariant it will send
the colimit diagrams to something we hope it to satisfy the universal
property of a limit in our target category. Now this can be seen as (at
least it's how I see it when I learned it some week ago, and I'm just a
student myself so take this lightly) that a presheaf is a sheaf if it
preserves gluing structures -- just like a group homomorphism preserves
the group structures!

Lukas Horosiewicz

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Dec 14, 2005, 6:14:34 AM12/14/05
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I tried to read this again but I didn't understand what you were trying
to say. If you haven't already show that the equalizer of a diagram of
say AbGrp A => B (f,g) is just {a \in A | f(a) = g(a)} (the universal
property gives that we only can get this up to isomoprhism but that's
good enough to consider that group above as a representative for the
equalizer), I think you got some of the properties messed up.
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