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

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?

• $\Gamma_F$ has no morphisms $(0,A)\to(0,B)$ since there are no morphisms $A\to B$ Commented Aug 15, 2023 at 12:46
• @FShrike Thank you for your reply. I made a mistake. I wanted to define A as $\{0, 1, 2\}$ and B as $\{0, 1\}$. Commented Aug 15, 2023 at 12:50
• I fixed that mistake. Commented Aug 15, 2023 at 12:51
• @AlexKruckman Thank you for your comment. Yes, I forgot to write it. I'll add it. Commented Aug 15, 2023 at 14:23

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 $$g\circ f=h\circ f$$.

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).