Re: Sheaves and the empty set

On May 19, 12:47 am, Kira Yamato <kira...@xxxxxxxxxxxxx> wrote:
But I'm defining F(emptyset) = Z. Therefore, the sheafification
procedure should keep for all U
F(U) = Z.
Then it should be fine even if there are disjoint open sets.

I do realize the problem when trying to speak of sheaves and when one
knows of the naive definition of sheaves of rings. Therefore some
authors define sheaves requiring F(emptyset)=0 the zero ring/group.
Some even define this at once on presheaves. But if you know the more
rigorous definition using products and so on, you won't face this
problem. An ellaborate explanation on this using the "naive"
definition but with more details on the covering (i.e. the author
tries to define "coverings" very precisely) can be found in the
lecture notes of Daniel Murfet (google his name).

Right sorry, its fine if there are -nonempty- disjoint open sets, but
when you deal with the empty set you get the problem.

To try to mend this problem when I say a family of f_i in F(U_i) have
the gluing property over a covering {U_i}_i\in I of U with U_i\subset
U, I mean that f_i|U_ij = f_j| U_ij for all i,j in I with U_ij=U_i/
\U_j. And when I say that these families can be glued together I would
mean that there exists a *unique* (which need not be said, because it
follows from the first sheaf axiom in the naive definition) f in F(U)
such that f|U_i = f_i for all i in I. Now I can take U_i=emptyset ,
and I=emptyset, I still have a covering of the emptset. And trivially
*any* family of f in F(emptyset) has the gluing property

remember

f_i|U_ij=f_j|U_ij for all ***i,j in I ****

the above condition is fulfilled for any family of f in F(emptyset)
because there is no i,j in I , the above becomes an empty condition.
Now what happens is that there exists indeed an f in F(emptyset) such
that

f|U_i = f_i for all *** i in I *** (1)

but the first sheaf axiom in the naive definition is now violated..
the first sheaf axiom says that

f|U_i = g|U_i for all ***i in I*** THEN f=g (2)

clearly any f satisfies (1) because we have an empty covering,
but then (2) is violated, because this f is not unique anymore.


I think the explanation above is the best one can do when one has this
naive definition.
An easier explanation is when you have the equalizer (or the exact
sequence as you call it).. If you have the empty covering the product
becomes empty, and empty product is the zero ring.

Well, I hope this elaborate explanation has helped you a bit.. but now
for the sheafification...

F in your definition is a presheaf there is no question to that
(unless one defines presheaf with F(emptyset)=0 ). But you are
realizing the sheafification incorrectly. The sheafification of F is
the sheaf induced by the family of rings F'(U), where

F'(U):={f:U --> Z : f locally constant}

at emptyset F'(emptyset) consists trivially of one point. Note that
the presheaf morphism between a presheaf and the sheaf need not be
monomorphism at each section (i.e. there are U by which
F(U)->F'(U) is not a monomorphism).

But the definition calls for any *arbitrary* open covering, not just a
particular open covering as you have chosen with the product set with
the empty indexing set. So you should take the maximum covering which
would include the emptyset, hence the covering would be nonempty.


Yes, but when you want to prove that something is NOT a sheaf, then
you choose *some* open covering and family of f_i's by which you
cannot glue.

--

-kira


Sincerely,
Jose Capco

.

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