Timeline for A simple(possibly trivial) question about Grothendieck Topologies
Current License: CC BY-SA 2.5
9 events
when toggle format | what | by | license | comment | |
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Feb 8, 2011 at 13:27 | comment | added | Allen Knutson | Also, if you want $\prod (A \coprod B) \cong \prod A \times \prod B$, you'd better take $\prod \emptyset$ to have one element. | |
Feb 8, 2011 at 1:39 | comment | added | Tom Leinster | Right, the key is that the empty product of sets is the 1-element set. Intuitively, this is for the same reasons that the empty product of numbers is 1, or the empty sum of numbers is 0. Rigorously, the categorical definition of product implies that an empty product is precisely a terminal object; and the terminal object of the category of sets is the 1-element set. | |
Feb 7, 2011 at 22:09 | comment | added | Rex | As $\Pi F(U_i\times U_j)$ is also just a single point, applying the equalizer condition gives that $F(\phi)$ is exactly one point. Thanks. | |
Feb 7, 2011 at 22:04 | comment | added | Rex | Ah!, I didn't know that the empty product is taken to be the 1-element set. I am reading some notes by Artin and there he says that every such functor is representable and that $F(U)=Hom(U,F(e))$. Thus, $F(\phi)$=1-element set. So how to show that $F(\phi)$ cannot be empty? Thanks. | |
Feb 7, 2011 at 21:56 | comment | added | darij grinberg | Read $\emptyset$ for $0$, sorry. | |
Feb 7, 2011 at 21:56 | comment | added | darij grinberg | ... can't have more than one element. | |
Feb 7, 2011 at 21:56 | comment | added | darij grinberg | I think the definition of a "sheaf of sets" includes the axiom that for every covering $\left\{U_i\right\}$ of an open set $U$, the map $F\left(U\right)\to\prod\limits_i F\left(U_i\right)$ is injective, and (this is something I am not 100% sure about, but I can't believe that Grothendieck would define otherwise) this includes the case when $U=0$ and the covering consists of zero sets. In this case, of course, $\prod\limits_i F\left(U_i\right)=\text{ empty product }=\text{ 1-element set}$, so $F\left(\emptyset\right)\to\text{ 1-element set}$ must be injective, so $F\left(\emptyset\right)$ ... | |
Feb 7, 2011 at 21:51 | history | edited | darij grinberg | CC BY-SA 2.5 |
no need for \phi when there is \emptyset
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Feb 7, 2011 at 21:46 | history | asked | Rex | CC BY-SA 2.5 |