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I was wondering if an equivalence of categories preserves group objects?

Intuitively yes but I can't show it.

given a group object $C = (C,\mu,\nu,\sigma)$ (respectively multiplication, unit and inverse) of a category $\cal{C}$, and $\cal{D}$ an equivalent category ($F:{\cal{C}\rightarrow D},G:\cal{D}\rightarrow C$) is $FC$ a group object?

I would like to say that I define $\mu' := F\mu(G\times G)(FC\times FC)$ intuitively this makes sense but $GFC \neq C$. What else could be done?

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    $\begingroup$ An equivalence of categories preserves limits and commutative diagrams: as these are the only things you need to define a group object I'd say your hypothesis is true $\endgroup$ – TheMadcapLaughs Dec 27 '17 at 14:36
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Here's the short way to show this. Assume that categories $\mathcal{C}$ and $\mathcal{D}$ have finite products, and let $F : \mathcal C \to \mathcal D$ be any functor that preserves finite products. Since the theory of groups is a Lawvere theory, a group object in a category $\mathcal{E}$ with finite products is the same thing as a finite product preserving functor $\mathcal{T}_{\mathbf{Grp}}\to\mathcal{E}$ where $\mathcal{T}_{\mathbf{Grp}}$ is the Lawvere theory of groups. If $H:\mathcal{T}_{\mathbf{Grp}}\to\mathcal{C}$ is a group object, then so is $F\circ H : \mathcal{T}_{\mathbf{Grp}}\to\mathcal{D}$ since the composition of finite product preserving functors is finite product preserving.

That an equivalence of categories preserves limits (and colimits) and thus finite products in particular, is easiest and most useful to see by rectifying it to an adjoint equivalence at which point it becomes a standard fact about adjoint functors.

If you don't want to assume that all finite products exist in $\mathcal{C}$ and $\mathcal{D}$, then you will still need to assume that in $\mathcal{C}$ all representatives of $C^n$ for each $n$ exists for the notion of group object (on $C$) to make sense. This will imply that the finite product $(FC)^n$ for each $n$ exists in $\mathcal{D}$. I say "representatives" because finite products (and limits/colimits in general) are only determined up to unique isomorphism and so there are many objects that can claim to be, e.g. $C^2$, in particular, both $F(C\times C)$ and $FC\times FC$ can claim to be $(FC)^2$. You can then apply the earlier argument to the full subcategories of $\mathcal{C}$ and $\mathcal{D}$ with finite products generated by $C$ and $FC$ respectively.

Going back to assuming all finite products, there is a natural isomorphism $\varphi: F(A\times B)\to FA\times FB$ natural in $A$ and $B$ and an isomorphism $\psi : F1 \to 1$ witnessing $F$ preserving finite products. Concretely, given a group object $(C,\mu,\nu,\sigma)$, you get a new group object $(FC, F\mu\circ\varphi_{C,C}^{-1},F\nu\circ\psi^{-1},F\sigma)$. Showing that this satisfies the laws of a group object is mostly an exercise in applying the naturality of $\varphi^{-1}$.

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    $\begingroup$ Actually it's possible to define what it means for $G$ to be a group object without any of the products $G^n, n \ge 2$ existing; giving $G$ the structure of a group object means giving $\text{Hom}(-, G)$ the structure of a group object in the category of presheaves. It's an exercise using the Yoneda lemma to show that this is equivalent to the usual definition when the products $G^0, G^2$ and $G^3$ exist. A simple example where you need this is in the category of finite sets of at most some cardinality. $\endgroup$ – Qiaochu Yuan Dec 28 '17 at 9:49
  • $\begingroup$ I have to think about how I feel about that. I'm not at all denying it, and it's completely clear that you only need triples to articulate the group laws, but it seems weird to say that $\forall a,b,c,d.abcd=dcba$ is a nonsense question e.g. when $G^4$ doesn't exist. $\endgroup$ – Derek Elkins Dec 28 '17 at 9:59
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    $\begingroup$ It's not a nonsense question, it can be asked in the presheaf category even if $G^4$ doesn't exist. $\endgroup$ – Qiaochu Yuan Dec 28 '17 at 10:03
  • $\begingroup$ Sure, in the presheaf category you have all limits and everything is fine. (It might be more parsimonious and a bit clearer to consider just a finite-product completion. Of course, there's benefit to simply reusing theorems about presheaf categories.) To me this comes across as an external view that is only internalizable in the case that the category does in fact have the relevant finite products. $\endgroup$ – Derek Elkins Dec 28 '17 at 10:19
  • $\begingroup$ Why? The thing you want to do is multiply four different "elements" of $G$, say four different morphisms $1 \to G$ although we could use any object in place of $1$. You can always do this as follows: consider these four elements as together specifying an element of $\text{Hom}(1, G)^4$, and since $\text{Hom}(1, G)$ has a group structure I can meaningfully apply the group multiplication here to get an element of $\text{Hom}(1, G)$ as desired. I don't see this as particularly "external" to the base category. Here is another way to say it: $G$ is a group object iff... $\endgroup$ – Qiaochu Yuan Dec 28 '17 at 10:23

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