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I tried to convince myself that the two caracterizations of a presheaf that is a sheaf given in wikipedia are equivalent but I couldn't.

(F presheaf and notations from wiki) Let's take a simple case, the following inclusions of open sets: $$ U_1 \cap U_2 \subset U_i \subset U_1\cup U_2 =:U ,\quad (i=1,2)$$

The restriction maps $res_{U_i, U}:F(U)\rightarrow F(U_i) $ define the first arrow $F(U)\rightarrow F(U_1)\times F(U_2)$ in the equalizer diagram (cf. wikipedia link). (by def. of a product).

Very very explicitely, the two other maps $F(U_1)\times F(U_2)\rightarrow F(U_1)\times F(U_1\cap U_2)\times F(U_2\cap U_1)\times F(U_2) $ are defined by the following maps

  • $F(U_1)\times F(U_2)\overset{\pi_1}{\rightarrow} F(U_1) \overset{res_{U_1,U_1}}{\longrightarrow} F(U_1)\ $ ( $\pi_1$ canonical projection) $F(U_1)\times F(U_2)\overset{\pi_1}{\rightarrow} F(U_1) \overset{res_{U_1\cap U_2,U_1}}{\longrightarrow} F(U_1\cap U_2) $ $F(U_1)\times F(U_2)\overset{\pi_2}{\rightarrow} F(U_2) \overset{res_{U_1\cap U_2,U_2}}{\longrightarrow} F(U_1\cap U_2)\ $
    $F(U_1)\times F(U_2)\overset{\pi_2}{\rightarrow} F(U_2) \overset{res_{U_2,U_2}}{\longrightarrow} F(U_2)\ $

and

  • $F(U_1)\times F(U_2)\overset{\pi_1}{\rightarrow} F(U_1) \overset{res_{U_1,U_1}}{\longrightarrow} F(U_1)\ $ ( same as above)
    $F(U_1)\times F(U_2)\overset{\pi_2}{\rightarrow} F(U_2) \overset{res_{U_1\cap U_2,U_2}}{\longrightarrow} F(U_1\cap U_2)\ $ (order changed) $F(U_1)\times F(U_2)\overset{\pi_1}{\rightarrow} F(U_1) \overset{res_{U_1\cap U_2,U_1}}{\longrightarrow} F(U_1\cap U_2) $ (order changed) $F(U_1)\times F(U_2)\overset{\pi_2}{\rightarrow} F(U_2) \overset{res_{U_2,U_2}}{\longrightarrow} F(U_2)\ $ (same as above)

All in all, that equalizer condition is just saying that $$ res_{U_1\cap U_2, U_1} \circ res_{U_1,U} \overset{!}{=} res_{U_1\cap U_2, U_2} \circ res_{U_2,U}$$ which already holds because both equal $res_{U_1\cap U_2, U} $ from the def. of presheaf. This doesn't look like gluing. What did I get wrong?

Second question: the gluing axiom itself is said to be formulated for a concrete category such that sthg, whereas the equalizer condition holds for a category with products. Is one formulation more general than the other?

Community
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Noix07
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  • Rather than referring to Wikipedia, you should just write down the equivalent conditions. From your question it is not clear which to equivalent conditions are you trying to prove. – Babai Jul 27 '14 at 09:25
  • The link is already pointing to the part of the article in wikipedia where the conditions are given, writting everything again will just make things longer and more boring...

    But I start thinking that the equalizer condition is rather saying sthg about the object than the arrow

     – Noix07 Jul 27 '14 at 13:22

1 Answers1

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I haven't worked through all your notation, but I expect this is what you're looking for:

The equalizer you're looking for is $$E=\lbrace (x,y)\in F(U_1)\times F(U_2)|res_{U1,U12}(x)=res_{U2,U21}(y) \rbrace$$

There is a map from $F(U)\rightarrow E$ given by $t\mapsto(res_{U,U_1}(t),res_{U,U_2}(t))$.

There is no reason to expect this map to be either injective or surjective. The sheaf condition says that it is both.

WillO
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  • (not a comment, but just as a note I can find later) $$ F(U) \overset{f}{\longrightarrow}\prod_{i\in I} F(U_i) {\overset{g}{\longrightarrow}\atop \underset{h}{\longrightarrow} } \prod_{i,j\in I} F(U_i\cap U_j)$$ $$ F(U) \overset{\cong}{\longrightarrow} E {\overset{g}{\longrightarrow}\atop \underset{h}{\longrightarrow} } \prod_{i,j\in I} F(U_i\cap U_j)$$ – Noix07 Apr 05 '15 at 14:20