In his descent II Bourbaki paper, Grothendieck lays out some criteria for a functor $F:C\rightarrow\mathrm{Set}$ to be strictly pro-representable; i.e. isomorphic to an inductive limit of representable functors $Hom(X_i,-)$ where X_i form a projective system with transition morphisms which are epimorphims. Assume $C$ has finite projective limits.

Then Grothendieck forms the comma category $CC=(1\downarrow F)$ and makes the following definitions. Object $A$ of $CC$ dominates object $B$ of $CC$ if there is an arrow $A\rightarrow B$ in $CC$, An object $A$ of $CC$ is minimal if whenever it is dominated by some $B$, such that the morphism in $CC$ is induced by a strict monomorphism $f$ in $C$, $f$ is an isomorphism. He then makes the following claims for $A,B$ in $CC$.

1) If $A$ is minimal and dominates $B$ by an arrow induced by an arrow $v$ in $C$, and $F$ is left exact (commutes with finite projective limits) then $v$ is unique. 2) if $B$ is minimal and dominated by $A$ then $v$ is epi. 3) $F$ is strictly pro-representable iff it is left exact and every object in $CC$ is dominated by a minimal object.

I don't see these things basically because I don't know how to cook up strict monomorphisms to use the minimality hypothesis. For (3) I'm aware of how the proof that any functor is a colimit of representable functors uses the same comma category. But still I need pointers…


1 Answer 1


If I try to prove 1:

Take objects $A,B$ given by a morphism $f:x\rightarrow y$ in $C$ and objects $u\in F(x)$, $v\in F(y)$ such that $v=F(f)(u)$. Consider another morphism $g:x\rightarrow y$ such that $v=F(g)(u)$. Then take $h:z\rightarrow x$ to be the equalizer of $f,g$. If I understand correctly $h$ is a typical example of a strict monomorphism. And since $F$ is left exact, $F(h):F(z)\rightarrow F(x)$ is the equalizer of the two maps $F(f), F(g): F(x) \rightarrow F(y)$. Of course $u\in F(z)$ since $F(f)(u) = F(g)(u)$. Then the minimality axiom implies that $h$ is an isomorphism, to that $f=g$.

I have no proof for 2 for the moment but this should not be more difficult.

To prove 3:

It's just the usual proof that $F$ is canonically a colimit of representable functors, but using only minimal objects. Then by 2 the transition morphisms will automatically be epimorphism, and you construct a strict pro-object. The usual proof goes as follows: there is a canonical map $$\operatorname{colim}_{(x,u)\in 1\downarrow F}\quad \hom(x,-) \longrightarrow F $$ basically because by Yoneda lemma a morphism $\hom(x,-)\rightarrow F$ is defined uniquely by an object of $F(x)$, corresponding to the image of $1_x \in \hom(x,x)$. This canonical map is always a injective (since you take the colimit over all these couples). In the classical case where you consider all of the comma category it is also surjective for more or less tautological reasons. Here the hypothesis tells you can restrict to minimal couples in the comma category and this will again be surjective.

  • $\begingroup$ thank you Louis-Clément. I was missing the point that any equalizer is a strict monomorphism. with $F$ still left-exact then a pf for (2): let $u:A\rightarrow B$ with $B$ minimal and $\xi\in F(A)$, $\xi'=F(u)(\xi)\in F(B)$, $v_1,v_2:B\rightarrow Z$ for $Z\in C$ such that $v_1\circ u=v_2\circ u$; we need to prove $v_1=v_2$. Form the equalizer $e:E\rightarrow B$ of $v_1$ and $v_2$; it is a strict monomorphism. Let $\xi''=F(v_1)(\xi')$, then $\xi''=F(v_2)(\xi')$ as well since $\xi'=F(u)(\xi)$. So $\xi'\in F(E)$ since $F(e)$ equalizes $F(v_1)$ and $F(v_2)$. $B$ is dominated by $e$ so $e$ Is an iso $\endgroup$
    – Tomo
    Nov 30, 2017 at 3:41
  • $\begingroup$ Yes it seems perfectly correct like this ! $\endgroup$ Dec 1, 2017 at 13:26
  • $\begingroup$ Pardonnez-moi de vous imposer une fois de plus mais il y a une chose de plus qui m'échappe… Grothendieck dit qu'on peut prendre la catégorie d'index I d'être ordonnée filtrante, et de plus il dit qu'on peut prendre le système projectif de façon que 'tout épimorphisme soit équivalent à un morphisme X_i -> X_j'… savez-vous ce qu'il veut dire par ça ? Et connaissez-vous où dans la littérature on peut trouver de tels résultats ? Merci bien ! $\endgroup$
    – Tomo
    Dec 2, 2017 at 0:33
  • $\begingroup$ Pour prendre I ordonné : il suffit d'écrire, étant donnés $f,g:i\rightarrow j$, correspondant à deux morphismes $X(f), X(g):X_j \rightarrow X_i$, on prend $h:j\rightarrow k$ tel que $hf=hg$ alors $X(f)\circ X(h)=X(g)\circ X(h)$ donc (comme $X(h)$ est supposé être un épimorphisme) $X(f)=X(g)$. Donc on peut supposer qu'il y a un unique morphisme $i \rightarrow j$, et donc que $I$ est ordonné filtrant. Pour la suite je ne comprends pas bien moi non-plus.. et je n'ai que l'article de Grothendieck comme référence. $\endgroup$ Dec 3, 2017 at 16:59
  • $\begingroup$ Thanks – after posting my question I also found this general result of Deligne in SGA IV_I: ‘Proposition 8.1.6. — Soit I une petite catégorie filtrante. Alors il existe un petit ensemble ordonné E, et un foncteur cofinal ϕ : EE → I, où EE désigne la catégorie associée a E.’ I still don't know about my other question. $\endgroup$
    – Tomo
    Dec 16, 2017 at 20:51

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