Wikipedia give sheaf property using equalizer diagram by saying sheaf property means for any open cover $\{U_i\}$ of $U$

$$F(U) \rightarrow \prod_{i} F(U_i) {{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} \prod_{i, j} F(U_i \cap U_j)$$

is an equalizer. What means two arrows? If $i = 1,2,3$ then what means

$$F(U) \rightarrow F(U_1) \times F(U_2) \times F(U_3) {{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} F(U_1 \cap U_2) \times F(U_2 \cap U_3) \times F(U_1 \cap U_3)?$$

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    $\begingroup$ There are textual descriptions above and below the indicated diagram at the cited Wikipedia article. Have you read those? $\endgroup$ – Zhen Lin Jul 27 '13 at 1:44
  • $\begingroup$ @ZhenLin I read it but I still don't understand. I am confused what is meant there. $\endgroup$ – user23086 Jul 27 '13 at 2:14
  • $\begingroup$ Is there a typo when you say: Similarly , fo each of $i,j \in I$, we have a map $b_{i,j}$: $$res^{U_j}_{U_i \cap U_j} \circ p_j : \prod_{i \in I} F(U_i) \rightarrow F(U_i \cap U_j)$$. Shouldn't it be: $$res^{U_j}_{U_i \cap U_j} \circ p_j : \prod_{i \in I} F(U_j) \rightarrow F(U_i \cap U_j)$$ because you are restricting $U_j$ to $U_i \cap U_j$ unless I am not understanding something. Sorry i couldnt put that in comments because my reputation is not 50 above!! $\endgroup$ – Avadutta Jun 12 '18 at 6:04

Given maps $f:X\to Y$ and $g:X\to Y$, a natural way of writing them together in the same diagram would be like this: $$X{{f\atop {\Large\longrightarrow}}\atop{{\Large\longrightarrow}\atop g}}Y$$ A map $\mathrm{eq}:E\to X$ is an equalizer (Wikipedia) of the maps $f$ and $g$ if it is final in the category of maps to $X$ that equalize $f$ and $g$. Depicting it together with the maps $f$ and $g$, we say that $$E\xrightarrow{\;\mathrm{eq}\;} X{{f\atop {\Large\longrightarrow}}\atop{{\Large\longrightarrow}\atop g}}Y$$ is an equalizer diagram.

Let $X$ be a topological space, let $F$ be a presheaf on $X$, and let $\{U_i\}_{i\in I}$ be an open cover of $X$. The presheaf $F$ comes with restriction maps $\mathrm{res}_{V}^{U}:F(U)\to F(V)$ for each pair of open sets $U,V$ with $V\subseteq U$. In particular, for each pair of $i,j\in I$, we have restriction maps $$\large\mathrm{res}_{U_i\cap U_j}^{U_i}:F(U_i)\to F(U_i\cap U_j)$$ and $$\large\mathrm{res}_{U_i\cap U_j}^{U_j}:F(U_j)\to F(U_i\cap U_j).$$ By definition, for each $i\in I$, the product (Wikipedia) $\prod_{i\in I}F(U_i)$ comes with a projection map $$p_i:\prod_{i\in I}F(U_i)\longrightarrow F(U_i).$$ Moreover, $F(U_i)$ comes with a restriction map $\mathrm{res}_{U_i\cap U_j}^{U_i}:F(U_i)\to F(U_i\cap U_j)$ for each $j\in I$. Composing them, we have for each pair of $i,j\in I$, a map $a_{i,j}$, as follows: $$\underbrace{(\mathrm{res}_{U_i\cap U_j}^{U_i}\circ p_i)}_{\Large a_{i,j}}:\prod_{i\in I}F(U_i)\longrightarrow F(U_i\cap U_j)$$ By the definition of a product, these $a_{i,j}$'s induce a map (let's call it $\alpha$) from $\prod_{i\in I}F(U_i)$ to the product of all of the $F(U_i\cap U_j)$'s together: $$\alpha:\prod_{i\in I}F(U_i)\longrightarrow \prod_{i,j\in I}F(U_i\cap U_j).$$ Similarly, for each pair of $i,j\in I$, we have a map $b_{i,j}$: $$\underbrace{(\mathrm{res}_{U_i\cap U_j}^{U_j}\circ p_j)}_{\Large b_{i,j}}:\prod_{i\in I}F(U_i)\longrightarrow F(U_i\cap U_j)$$ which come together to form a single map $$\beta:\prod_{i\in I}F(U_i)\longrightarrow \prod_{i,j\in I}F(U_i\cap U_j).$$ These maps $\alpha$ and $\beta$ are the top and bottom arrows in the diagram $$F(U) \longrightarrow \prod_{i\in I} F(U_i) {{{}\atop {\Large\longrightarrow}}\atop{{\Large\longrightarrow}\atop {}}}\prod_{i, j\in I} F(U_i \cap U_j)$$ We say that $F$ is a sheaf (Wikipedia) if, for any open cover $\{U_i\}_{i\in I}$ of our space $X$, this diagram is an equalizer diagram.

Note that $\alpha$ and $\beta$ are not the same map; they take different "paths". Let's trace out what happens: the top-right path is $a_{i,j}$, and the bottom-left path is $b_{i,j}$. $\require{AMScd}$ $$\require{AMScd} \begin{CD} \prod_{i\in I}F(U_i) @>{\Large p_i}>> F(U_i);\\ @V{\Large p_j}VV @VV{\Large\mathrm{res}_{U_i\cap U_j}^{U_i}}V \\ F(U_j) @>>{\Large\mathrm{res}_{U_i\cap U_j}^{U_j}}> F(U_i\cap U_j); \end{CD}$$ The two maps $\alpha$ and $\beta$ to the product $\prod_{i,j\in I}F(U_i\cap U_j)$ agree on a specific element $(s_i)_{i\in I}$ of the product $\prod_{i\in I}F(U_i)$ if and only if $a_{i,j}$ and $b_{i,j}$ agree on it for all pairs of $i,j\in I$. But $a_{i,j}$ and $b_{i,j}$ are clearly going to be different maps in general, so this is not a trivial condition.

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    $\begingroup$ @Zee: The comment below is wrong, the product $$\prod_{i\in I}F(U_j)$$ is just $F(U_j)$ to the power $|I|$ (the cardinality of $I$), a very different object than $$\prod_{i\in I}F(U_i)$$ $\endgroup$ – Zev Chonoles Apr 5 '19 at 14:22
  • $\begingroup$ It looks to me like one either has to distinguish between $F(U_i \cap U_j)$ and $F(U_j \cap U_i)$, or that one has to specify for each unordered pair $(i,j)$ which of the two maps $\prod_{i} F(U_i) \to F(U_i \cap U_j)$ one is choosing for $\alpha$ and which for $\beta$. $\endgroup$ – Improve May 13 '19 at 22:45

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