Re: Sheaves and the empty set

On 2007-05-18 17:29:40 -0400, Jose Capco <cliomseerg@xxxxxxxxxxxxxxxxxxxxxxxxx> said:

On May 18, 9:49 pm, Kira Yamato <kira...@xxxxxxxxxxxxx> wrote:

Oh, is the requirement that
F(emptyset) = zero ring
be part of the definition for a sheaf? Otherwise, it seems that my
example satisfies the definition for a sheaf.

No, it's not part of the definition. It can be deduced. In your
example you said for *all* open set U.. did you meant that
F(emptyset)=0 ..

I meant for every open sets U, including the emptyset, in X
F(U) = Z.

besides what are you restriction maps?

If V is an open subset of an open set U, then the restriction map
F(U) = Z --> Z = F(V)
is always the identity map on Z. This includes the case when U=V.

If your
restriction maps is the identity for nontrivial restriction and if
F(emptyset)=0 your example wouldn't work if your topological space had
two disjoint open set.

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.



About your exact sequence

F(U) --> \prod_i F(U_i) ==> \prod_i, \prod_j F(U_i \cap U_j)

isn't the emptyset a covering (of a single open set) of the emptyset
itself? So, the products in the exact sequence are not empty products.


it is , but I didnt take the emptyset as my covering.. I took the
empty covering (which is something else ;) )..

What I took is \/_{i\in\emptyset} U_i = emptyset
Which forces the product to be an empty product, making in the zero
ring.

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.

--

-kira

.

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