# Are poset-shaped limits of finite groups profinite?

My basic question is

Question 0: Let $\mathcal{C}$ be a complete category and $\mathcal{D}$ its full subcategory such that $\mathcal{D}$ is closed under finite limits (in $\mathcal{C}$). Let $I$ be a small thin category and $\Delta: I \rightarrow \mathcal{D}$ a diagram. Is $X:=\lim_{\mathcal{C}} \Delta$ an inverse limit? That is, does there exist a downwards directed poset $J$ and a diagram $\Delta^*: J \rightarrow \mathcal{D}$ such that $X=\lim_{\mathcal{C}} \Delta^*$?

It seems to me that the answer is yes: One can define objects of $J$ to be finite subsets $I_0$ of $Ob(I)$. Order these by reverse inclusion $"\leq"$, those are the morphisms. The set $(J, \leq)$ is then obviously directed downwards. Given $I_0 \in Ob(J),$ set $\Delta^*(I_0)=\lim_\mathcal{C}(\Delta \restriction_{I_0})$, where $\Delta \restriction_{I_0}$ is $\Delta$ precomposed with the inlucion of the full subcategory on $I_0$ to $I$. This belongs to $\mathcal{D}$ by assumptions. Given $I_0 \subseteq I_1 \in Ob(J),$ the morphism $\Delta^*(I_1) \rightarrow \Delta^*(I_0)$ is induced from the cone of $\Delta^*(I_1)$ over $\Delta \restriction_{I_0}$, using the universal property of $\Delta^*(I_0)$ .

So basically this construction throws in limits of finite subdiagrams to make the diagram directed downwards. It seems obvious that this does not change the limit, meaning: the inclusion of categories $I \subseteq J$ (objects to singletons) induces a unique isomorphism $\lim_{\mathcal{C}} \Delta^* \rightarrow \lim_{\mathcal{C}} \Delta$ (compatible with the obvious cones).

Question 1: Does this really work, or am I neglecting something?

As an example of this, setting $\mathcal{C}$ to be the cat. of topological groups and $\mathcal{D}$ the full subcat. of finite discrete groups, one would obtain that profinite groups can be equivalently defined as limits of finite groups with respect to general posets, not necessarily directed ones. In particular, a product $G=\prod_{i \in I} G_i$ of finite discrete groups is always profinite, since it is Hausdorff, totally disconnected and compact. To realize it as an inverse limit, one way to do this is to write $G=\varprojlim_{I_0} \prod_{I_0}G_i$, which agrees with the above construction.

Question 2: If this works, how much can this be generalized? Does a version of this strategy work when $I$ is small and locally finite, i.e. hom-sets are finite? More generally, what is an example of a diagram $\Delta$ to discrete finite groups such that $\lim \Delta$ is not profinite?

Thanks very much for any help.

In detail, suppose $\mathcal{C}$ is any category with finite limits and $\Delta:I\to\mathcal{C}$ is a diagram. Let $J$ be the set of all pairs $(o,m)$ where $o$ is a finite set of objects of $I$ and $m$ is a finite set of morphisms between objects of $o$. We order $J$ by reverse inclusion on each coordinate, so $J$ is directed downwards. For each $j=(o,m)\in J$, let $L_j$ be the limit of the diagram formed by $j$. Explicitly, consider the category $F_j$ whose objects are $o\times\{0,1\}$ and has a morphism $(A,0)\to (B,1)$ for each morphism $A\to B$ in $m$. There is an obvious functor $F_j\to I$ and $L_j$ is the limit of the composition $F_j\to I\stackrel{\Delta}{\to}\mathcal{C}$. (Note that we use this construction $F_j$ instead of just the subcategory of $\Delta$ generated by $m$ since that subcategory may not be finite, while $F_j$ always is finite since we have "forgotten" about composability of our morphisms.)
These finite limits $L_j$ then naturally form a diagram indexed by $J$. An inverse limit of this diagram can easily be verified to be the same thing as a limit of the original diagram $\Delta$.