Is a category of elements of a functor $F \colon C \to Set$ where $C$ has only epis really a poset?

Question

I'm reading On the Galois Theory of Grothendieck, and I came across an argument I suspect is wrong. Please see the beginning of Ⅲ in the paper.

First, let $\mathcal{C}$ be a category with only epis, $F$ a functor $\mathcal{C} \to \mathsf{Set}$, and $\Gamma_F$ the category of elements of $F$. 3.2 says that such $\Gamma_F$ is a poset (more strongly, strict-epicness is supposed, but 3.2 doesn't use it). $\Gamma_F$ can have two objects isomorphic to each other in the first place, so it should be a "proset" or "pre-ordered set". However, it is just a misuse of terms, so it's not important. But I believe that $\mathcal{C}$ is not always even a proset. For example, let $\mathcal{C}$ be a subcategory of $\mathsf{Set}$ which has only two objects, $A = \{0, 1, 2\}$ and $B = \{0, 1\}$, and all the surjections as morphisms, and $F$ the inclusion functor, then $\Gamma_F$ has three morphisms $(0, A) \to (0, B)$, and it's not a proset. I guess the problem is that the canonical maps of the colimit $[A, −] → F$ can be not injective because the diagram $(\Gamma_F)^{\mathbf{op}} \to \mathsf{Set}^{\mathcal{C}}$ is not always filtered. And if $(\Gamma_F)^{\mathsf{op}}$ is also filtered, I guess that all the canonical morphisms $[A, −] → F$ are injective and that $\Gamma_F$ is a proset.

Second, the axiom C3 says that $\Gamma_F$ has all finite meets, and it leads that $(\Gamma_F)^{\mathsf{op}}$ is filtered (in C3, it's not said what is filtered is $(\Gamma_F)^{\mathsf{op}}$, but I guess it should be $(\Gamma_F)^{\mathsf{op}}$ by 3.5). But the fact $\Gamma_F$ is a proset must depend on the fact $\Gamma_F$ is filtered.

Then, I think that the axioms in 3.1 are not enough, and it should be also supposed that $\Gamma_F$ is filtered. Or $\Gamma_F$ isn't a proset and it doesn't affect later arguments. (I don't know which is true.)

Is my suspicion true, and which above is true?

Answer 1

The functor $F\colon C\to\mathsf{Set}$ sending all objects to singletons has its category of elements $\Gamma_F$ isomorphic to $C$, preserves quotients, is valued in finite sets, and preserves strict epimorphisms.

Then C0), C1), and C2) are satisfied automatically if all morphisms of $C$ are epimorphisms and every object has a quotient by any action of a finite group, but $\Gamma_F$ is not a poset unless $C$ is.

Taking $C$ to be the group of integers $\mathbb Z$ considered as a one-object category gives such an example (all morphisms are isomorphisms, so strict epimorphisms, and the only actions by finite groups are trivial with the identity morphism as quotient). Moreover, $C$ is filtered in the sense appropriate for posets described in Remark 3.5: any two objects $A$ and $B$ admit a morphism with a common domain $C$.

However, $C\cong\Gamma_F$ does not have finite products (the analogue of finite meets for categories).

In fact, $\Gamma_F$ having finite products (C3 stated for categories rather than posets) and all morphisms of $C$ being epimorphisms suffices to show $\Gamma_F$ is a pre-ordered set. Moreover, the necessary and sufficient condition for $\Gamma_F$ to be a poset when all morphisms of $C$ are epimorphisms is the additional cofilteredness condition for categories: that for any two morphisms $g,h\colon Y\rightrightarrows Z$ there exists a morphism $f\colon X\to Y$ such that $g\circ f=h\circ f$.

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By definition, any fork $(a,A)\overset f\to (b,B)\overset{g_1}{\underset{g_2}\rightrightarrows}(c,C)$ in the category of elements $\Gamma_F$ of a functor $F\colon\mathcal C\to\mathsf{Set}$ determines a fork $A\overset f\to B\overset{g_1}{\underset{g_2}\rightrightarrows} C$ in $\mathcal C$. In particular, if $f\colon A\to B$ is an epimorphism, then $g_1=g_2$, and so $(a,A)\overset f\to(b,B)$ is also an epimorphism in $\Gamma_F$.

Thus if all morphisms in $\mathcal C$ are epimorphisms, then so are all morphisms in $\Gamma_F$.

Now in a category all morphisms of which are epimorphisms, if an object $Y$ of $\Gamma_F$ admits a square $Y\times Y$, then the diagonal $\Delta\colon Y\to Y\times Y$ is both a split monomorphism and an epimorphism, so is an isomorphism. It follows that any two morphisms $X\rightrightarrows Y$ are equal. Thus a category with squares and all of whose morphisms are epimorphisms has at most one morphism between any two objects, i.e. is a pre-order.

In particular, requiring that $\Gamma_F$ has finite products (the analogue of finite meets for categories) forces it to be a pre-order. Conversely, finite meets in a pre-order are the same as finite products.

In fact a category all of whose morphisms are epimorphisms is a pre-order if and only if or any pair of morphisms $g,h\colon Y\rightrightarrows Z$, there exists a morphism $f\colon X\to Y$ such that $g\circ f=h\circ f$.

Traditionally, the definition of cofilteredness includes both the above condition and that for any two objects $Y$ and $Z$ there is a pair of morphisms with common codomain $X\to Y$ and $X\to Z$. The latter condition is cofilteredness for partially ordered sets, but in a general category all of whose morphisms are epimorphisms does not suffice to conclude the category is a pre-order (e.g. the category $\bullet\rightrightarrows\bullet$ or the example above).