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I'm reading Theorem 2.9.1 of this notes about glueing sheaves which uses a categorical argument.

enter image description here Question: In (iii) below it's stated that there is an isomorphism $F_i(U_i\times _X V)\cong \underset{\leftarrow}{\operatorname{lim}}_{W\in \tau} F_i(U_i\times_X W)$ where I assume $\tau$ is a covering of $V$. How the limit diagram is constructed here? Does this limit contain the same information as the equaliser diagram in the definition of sheaves? enter image description here

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  • $\begingroup$ Have you looked at the isomorphism (2.6) ? $\endgroup$ – Maxime Ramzi Jan 16 '20 at 20:11
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The answer is yes, basically because all limit diagrams can actually be written as an equalizer of products.

Here $\mathcal T$ seems to be a typo for $\mathcal V$ which is a covering of $V$ assumed to be stable under fiber products. Then I claim that the limit of $F_i(U_i\times_X W)$ over $\mathcal V$ is the usual thing, that is : the equalizer of $\prod_{W\in \mathcal V}F_i(U_i\times_X W) \rightrightarrows \prod_{W,W'\in \mathcal V}F_i(U_i\times_X (W\times_V W'))$ where the two maps are the same as usual.

If this is correct, then the fact that the limit is $F_i(U_i\times_X V)$ is just the statement that $F_i$ is a sheaf and that $U_i\times_X W$ is a covering of $U_i\times_X V$ and that $U_i\times_X (W\times_V W') = (U_i\times_X W)\times_{U_i\times_X V} (U_i\times_X W')$ (which follows from some simple diagram chase)

Now the proof. Suppose you have a cone $(Y,f_W)$ over the $F_i(U_i\times_X W), W\in\mathcal V$. Then clearly the product map $Y\to \prod_{W\in \mathcal V}F_i(U_i\times_X W)$ with coordinates the $f_W$ equalizes the two maps of the above equalizer, as $W\times_V W' \in \mathcal V$.

So we get a unique map $Y\to$ the equalizer.

So it suffices to show that the projections of the equalizer form a cone. But suppose you have a map $W\to W'$ in $\mathcal V$, that is, an inclusion $W\subset W'$. Then $W'\times_V W = W$ so that clearly the two maps are equalized by the equalizer : it does form a cocone.

So we are done

(I treated the case of a space, but of course if there is a map $W\to W'$ in $\mathcal V$, then it's a map above $V$, so that $W\times_V W'$ is still canonically identified with $W$, so it doesn't change much)

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  • $\begingroup$ Thanks a lot! For $\mathcal V\in Cov(U)$ viewed as a category, is the limit indexed over the category $\mathcal V$ the equaliser in the sheaf condition? Is your proof showing the limit and the equaliser are the same? Why is it done by showing there is a unique map from a cone to the equalizer and the projection of the equalizer form a cone? $\endgroup$ – Alex Jan 17 '20 at 8:59
  • $\begingroup$ Yes, that's what I proved. Well the second part shows that the equalizer is a cone, and the first part that it is a limit cone, so the equalizer is the limit (for instance if you take directly the product instead of the equalizer, you do get a unique map from each cone that commutes with the projections; but the product itself is not a cone, so it's not a limit) $\endgroup$ – Maxime Ramzi Jan 17 '20 at 9:15

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