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I'm now studying the etale cohomology with the book 'Introduction to Etale Cohomology' by Tamme.

In the page 26 of the book, 'a family of effective epimorphisms' is introduced.

'A family $\{ U_{i} \rightarrow V \}$ is a family of effective epimorphisms if the diagram

$Hom(V,Z) \rightarrow \prod_{i} Hom(U_{i}, Z) \rightrightarrows \prod_{i,j}(U_{i}\times_{U}U_{j},Z)$ is exact for all the objects $Z$.

The question is 'in the condition, do we restrict the pairs $i,j$ to be distinct?'

I gave a thought on this question, but I'm not sure whether the two versions give us the equivalent results or not.

Thank you very much in advance.

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2 Answers 2

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That exactness conditions can be rephrased more explicitely as:

$$ Hom(V,Z) = \left\lbrace (v_i) \in \prod_i Hom(U_i,Z) \ \middle| \ \forall i,j,v_i \circ \pi_1 = v_j \circ \pi_2 \right\rbrace $$

where $\pi_1,\pi_2$ denotes the two projections $U_i \times_V U_j \rightrightarrows U_i,U_j$.

When you write it like this, the condition in the case $i=j$ is clearly vacuous when all the map $U_i \to V$ are monomorphisms. Indeed in this case $U_i \times_V U_j$ is justs the intersection of $U_i$ and $U_j$, so that $\pi_1=\pi_2$ when $i = j$. This case is very frequent, and you can very often restrict to it by considering the "image" of the $U_i$ in $V$.

But in some situation (for e.g. if you want to keep your objects $U_i$ be to in some specified site that do not admit image factorization like the étale site) it might not be the case. and in general you need the case $i=j$. Consider the case where the you only have a single map $U \to V$. Then the condition becomes

$$ Hom(V,Z) = \left\lbrace f \in Hom(U,Z) \ \middle| \ f\circ \pi_1 = f \circ \pi_2 \right\rbrace $$

where $\pi_1$ and $\pi_2$ are the two projections $U \times_V U \rightrightarrows U$.

You can think of $U \times_V U \rightrightarrows U$ as a map $U \times_V U \to U \times U$ which is a monomorphisms and corresponds to the "equivalence relation such that $V$ should be the quotient of $U$ by this relations". Or rephrased this as $V$ being the coequalizer ('in the category of sheaves') $U \times_V U \rightrightarrows U \to V$, i.e. $V = U /R$ where $R$ is the equivalence relation $U \times_V U$.

And a function from $V \to Z$ can be described as a function $U \to Z$ which is compatible to the equivalence relation $R$ such that $U/R \simeq V$.

Also note that in the general case (with several map) you can think of the general condition as being in two part: you have the condition for $i=j$ that assert that each maps $U_i \to Z$ factors through "the image $V_i$ of $U_i$ in $V$" (if this make sense) , and the condition for $i \neq j$ that implement the usual compatibility condition.

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    $\begingroup$ Wait, the two restrictions to $U_i\times_U U_i$ can still be different, as the two projections $U_i\times_U U_i\to U_i$ don't necessarily agree (only if the maps $U_i\to U$ are monomorphisms). This doesn't come up in the small site of a topological space, but definitely in the context of more general sites and etale cohomology. $\endgroup$
    – Achim Krause
    Commented Nov 30, 2020 at 13:48
  • $\begingroup$ @Simon Henry So the condition for i=j has its own meaning. As I'm not completely easy with category theory, I don't perfectly understand your explanation on the equivalence relation. Can you suggest some 'general' class of categories where your explanation is more explicit and complete? $\endgroup$
    – gualterio
    Commented Nov 30, 2020 at 14:34
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    $\begingroup$ You can simply take the category of Sets (i.e., $U_i$, $V$ and $Z$ are all just sets), that should make everything clear. The general picture as I describe it informally always make sense when you work in the topos of sheaves instead of the site, but as a topos is basically "a category that behave like the category of set" looking at what happen in the category of sets should be illuminating enough. $\endgroup$
    – Simon Henry
    Commented Nov 30, 2020 at 14:46
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Yes, you have to include the case $i=j$. Just look at what happens in the case of a single $U_1$, in order for this to boil down to the concept of a single effective epimorphism (introduced in the preceding paragraph in the book) you need to include $i=j=1$.

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  • $\begingroup$ I agree with you that single effective epimorphism should make sense at least for the third condition: an isomorphism is a covering. For family consisting of more than one epimorphism, do you think that allowing i, j to be equal change the theory? $\endgroup$
    – gualterio
    Commented Nov 30, 2020 at 14:13
  • $\begingroup$ Yes, it changes the notion: think about the case of two epimorphisms $A \to X$ and $B \to Y$, and the map $A,B \rightrightarrows X \coprod Y$. See the last remark in my answer. $\endgroup$
    – Simon Henry
    Commented Nov 30, 2020 at 14:15

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