Following Anton Deitmar, let $\mathcal B$ be an "$\mathbb F_1$-linear category" (Deitmar uses the term "Belian"); i.e., $\mathcal B$ is balanced, pointed, contains finite products, kernels, and cokernels, and has the property that every morphism with zero cokernel is an epimorphism. Let $\mathsf{Set}_*$ denote the category of pointed sets, and let $Lex(\mathcal B,\mathsf{Set}_*)$ denote the category of left-exact functors $F: \mathcal B\to\mathsf{Set}_*$ (that is, $F$ preserves finite limits).

I want to show that $Lex(\mathcal B,\mathsf{Set}_*)$ is also $\mathbb F_1$-linear if $\mathcal B$ is small; in particular, the properties that still concern me are that $Lex(\mathcal B,\mathsf{Set}_*)$ is balanced, and that every morphism with $0$ cokernel is an epimorphism. I know that $Lex(\mathcal B,\mathsf{Set}_*)$ is a reflective subcategory of $Fun(\mathcal B,\mathsf{Set}_*)$ and that finite products, kernels, and cokernels exist in $Lex(\mathcal B,\mathsf{Set}_*)$ (it's also clear that $Lex(\mathcal B,\mathsf{Set}_*)$ is pointed), and I know that $Fun(\mathcal B,\mathsf{Set}_*)$ is $\mathbb F_1$-linear. My hope was originally to prove that if a natural transformation $\eta : F\to G$ is mono/epi in $Lex(\mathcal B,\mathsf{Set}_*)$, then it is mono/epi in the full functor category, which would show that $Lex(\mathcal B,\mathsf{Set}_*)$ is balanced (and I had a similar plan to show that morphisms with $0$ cokernel were epi), although a proof of this is turning out to be quite elusive (and I'm no longer convinced that one exists). I was able to show that if a natural transformation is a monomorphism in $Lex(\mathcal B,\mathsf{Set}_*)$, it is a monomorphism in the full functor category - epimorphisms are the root of my problems. Then my question is really a few related questions (hopefully they're related enough to warrant posting all of them here):

  1. Are epimorphisms in $Lex(\mathcal B,\mathsf{Set}_*)$ also epimorphisms in $Fun(\mathcal B,\mathsf{Set}_*)$?
  2. If not, does anyone know of an alternate proof (or a resource where I could find one) that a bimorphism in $Lex(\mathcal B,\mathsf{Set}_*)$ is also an isomorphism?
  3. Is there any reasonable condition one can place on $\mathcal B$ to force epis in $Lex(\mathcal B,\mathsf{Set}_*)$ to remain epis in $Fun(\mathcal B,\mathsf{Set}_*)$ or to force bimorphisms in $Lex(\mathcal B,\mathsf{Set}_*)$ to be isomorphisms? (I doubt that this has an answer in the affirmative, as my experience indicates that many of the properties of a functor category depend on the codomain rather than the domain.)
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    $\begingroup$ When you say "I want to show" do you mean "I know this is true (e.g. because Deitmar states it somewhere) and I'm trying to reproduce the proof" or do you mean "I'd like this to be true and I'm trying to write down the proof"? $\endgroup$ – Qiaochu Yuan Jan 31 '14 at 1:40
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    $\begingroup$ @QiaochuYuan: I mean "I'd like this to be true." I'm working on extending Deitmar's results, so there's not a guarantee that these claims should be true (although I do have reason to believe that they should be). However, that's why I'm also interested in question 3; if the results aren't true with the assumptions already made, I am not opposed to adding some other (reasonable) assumptions about $\mathcal B$ to salvage the statement. $\endgroup$ – Stahl Jan 31 '14 at 1:49
  • $\begingroup$ I think (1) is unlikely. It is not true when, say, $\mathcal{B}^\mathrm{op}$ is the category of finitely presentable rings and we look at $\mathbf{Lex} (\mathcal{B}, \mathbf{Set})$. $\endgroup$ – Zhen Lin Jan 31 '14 at 8:40
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    $\begingroup$ I just realized my earlier comment was a bit ambiguous: I don't have reason to believe that all the subquestions have answers in the affirmative, but rather that I have reason to think that $Lex(\mathcal B,\mathsf{Set}_*)$ is $\mathbb F_1$-linear. $\endgroup$ – Stahl Jan 31 '14 at 17:04
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    $\begingroup$ @Joseph: the difference between this case and the abelian case is that the axioms defining an abelian category are self-dual and these axioms aren't. $B^{op}$ is much more natural; for starters the corresponding construction becomes covariant in $B$, and also $B$ naturally embeds into $\text{Lex}(B^{op}, \text{Set}_{\ast})$, but more importantly, under mild hypotheses this is the category of ind-objects in $B$ (and so there's more hope of it inheriting properties from $B$). $\endgroup$ – Qiaochu Yuan Feb 2 '14 at 3:04

Here is a counterexample (modulo a small statement I'm not sure how to prove). Building on the comments, let $B$ be (a skeleton of) the opposite of the category of at most countable groups. This satisfies all of the desired conditions. There is a functor $\text{Grp} \to \text{Lex}(B, \text{Set}_{\ast})$ sending a group $G$ to the representable functor $\text{Hom}(-, G)$ (which lands in $\text{Set}_{\ast}$ since $\text{Grp}$ has a zero object) which is full and faithful; you can see this by evaluating on the subcategory given by the free groups, which gives a right adjoint $\text{Lex}(B, \text{Set}_{\ast}) \to \text{Grp}$ (this is the statement I'm not sure how to prove). In particular, $\text{Grp}$ is a coreflective subcategory, so colimits in $\text{Grp}$ agree with those in $\text{Lex}(B, \text{Set}_{\ast})$.

But $\text{Grp}$ has morphisms with zero cokernel which are not epimorphisms: take any inclusion $f : H \to G$ of a proper subgroup $H$ into a group $G$ whose normal closure is all of $G$ (for example, take $G$ to be simple and $H$ to be proper and nontrivial). Hence the same is true of $\text{Lex}(B, \text{Set}_{\ast})$.

For the purposes of imitating the proof of Freyd-Mitchell I think you should be looking at $\text{Lex}(B, \text{Set}_{\ast})^{op}$ as I mentioned in the comments.

  • $\begingroup$ You've given me a good bit to think about; I'll probably come back and accept this later after some thought about the counterexample and the comments, and I might wind up posting another related question once I think about what exactly will be done to fix the issue. Thanks Qiaochu! $\endgroup$ – Stahl Feb 4 '14 at 1:32
  • $\begingroup$ @Joseph: no problem. For completeness sake, above I implicitly use the fact that $\text{Grp}$ can be presented as a certain category of functors; see qchu.wordpress.com/2013/06/09/operations-and-lawvere-theories for some details. $\endgroup$ – Qiaochu Yuan Feb 4 '14 at 1:33

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