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For example, given etale maps $p : X \to B$, $r : Y \to B$ and continuous maps $f, g : X \to Y$ such that $p = r \circ f = r \circ g$, the equalizer of $f$ and $g$ seems to be $E = \lbrace x \in X \mid f x = g x \rbrace$ with the inclusion map $e : E \to X$ and the etale map $p \circ e : E \to B$. This works because whenever $f$ and $g$ agree at a point, they agree on a neighborhood of the point, and so $E$ is an open subspace of $X$.

Similarly, if I am not mistaken, the coequalizer of $f$ and $g$ is again computed simply as the quotient $Q = Y/{\sim}$ where $\sim$ is the least equivalence relation satisfying $f x \sim g x$ for all $x \in X$. The equivalence relation $\sim$ is fiber-wise, i.e., it never relates things from different fibers. The canonical quotient map $q : Y \to Q$ is compatible with $r$ and so we get a map $t : Q \to B$, which turns out to be etale.

Is this correct? Where can I find a reference? Most textbooks that prove the category of etale maps over a given base to be a topos make a detour via the category of sheaves on $B$, which makes the calculations of limits and colimits indirect, as they pass through the equivalence between etale maps and sheaves.

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It follows from Theorem 2 of II.6 of Mac Lane-Mordijk that the inclusion functor from etale maps over X to bundles over X is left adjoint to taking the etale space of the sheaf of cross-sections. Thus the inclusion preserves all colimits and so colimits are fiber-wise. Exercise 6 of that chapter implies the inclusion preserves finite limits and so they are fiber-wise as well. I think it is not hard to see that an infinite fiber product of etale maps is not etale. This comes from the fact that basic neighborhoods in the product topology only see finitely many factors.

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Thanks, in the meanwhile (7 months) I computed these facts myself with bruce force. –  Andrej Bauer Dec 3 '11 at 10:01
    
Sorry, I just saw the question now :) –  Benjamin Steinberg Dec 3 '11 at 12:04
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