8
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$ Nov 30 '20 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
    Nov 30 '20 at 14:34
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 30 '20 at 14:46
7
$\begingroup$

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 $i=j=1$.

$\endgroup$
2
  • $\begingroup$ 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
    Nov 30 '20 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. $\endgroup$ Nov 30 '20 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.