inverse limit isomorphisms Let $(I,\leq )$, $(J,\leq )$ be partially ordered sets and $G_{ij}$ be a set of groups indexed by pairs $(i,j)\in I\times J$.The product of the partially ordered sets $(I,\leq )$, $(J,\leq )$ is $I\times J$ with the partial order $(i,j)\leq (i',j')$ iff $i\leq i'$ and $j\leq j'$. Suppose that for all $(i,j)\leq (i',j')$ we are given homomorphisms $f_{ij,i'j'}:G_{i'j'}\rightarrow G_{ij}$ satisfying the usual transitivity conditions. How to prove that there are canonical isomorphisms : $\varprojlim_{i}\varprojlim_{j}G_{ij}\sim \varprojlim_{(i,j)}G_{ij}\sim \varprojlim_{j}\varprojlim_{i}G_{ij}$?
Thank you.
 A: Let's begin by defining some stuff: 
$$\begin{cases} G_{\bullet,j}:=\varprojlim_i G_{i,j} \\[4pt] G_{i,\bullet}:=\varprojlim_j G_{i,j} \\[4pt] G_{\bullet,\bullet}=\varprojlim_{(i,j)}G_{i,j}\end{cases}$$
In order to define the double inverse limits we need the following types of transition morphisms:
$$\begin{cases} G_{i',\bullet}\to G_{i,\bullet} \\[4pt] G_{\bullet,j'}\to G_{\bullet,j} \end{cases}$$
To see these, organize the situation into a commutative diagram like so:
$\hskip 1.2in$ 
We don't have morphisms $G_{\bullet,j'}\to G_{\bullet,j}$ automatically, but what we do have are downward projections $G_{\bullet,j'}\to G_{\ell,j'}$ and rightward transitions $G_{\ell,j'}\to G_{\ell,j}$, which upon composition yields more morphisms $G_{\bullet,j'}\to G_{\ell,j}$ as $\ell$ varies. As such, by the UP (universal property) of $G_{\bullet,j}$, we get a unique morphism $G_{\bullet,j'}\to G_{\bullet,j}$ commuting with all of the relevant arrows down below.
Symmetrically, the same reasoning applies to obtain unique downward transitions $G_{i',\bullet}\to G_{i,\bullet}$ commuting with all of the relevant arrows. At this point we can define the two double inverse limits, $H=\varprojlim_j G_{\bullet,j}$ (horizontal) and $V=\varprojlim_i G_{i,\bullet}$ (vertical).
It suffices to show $H$ satisfies the same universal property as $G_{\bullet,\bullet}$, since then they will be $\cong$ up to unique isomorphism (subject to commuting with the blue transitions) and further $V$ will also be canonically $\cong$ by symmetrical reasoning. (See Thm 1.1.3 on pg 8 here.)
Let $X$ be an object and define morphisms $\phi_{i,j}:X\to G_{i,j}$. For fixed $j$ the collection $\{\phi_{i,j}:i\in I\}$ induces by UP a unique morphism $\phi_{\bullet,j}:X\to G_{\bullet,j}$ commuting with the family. Furthermore the family $\{\phi_{\bullet,j}:j\in J\}$ induces by UP a unique morphism $\phi:X\to H$ commuting with all of the induced maps $\phi_{\bullet,j}$ and hence with all of the original maps $\phi_{i,j}$ (by post-projecting down to the blue region). Thus we have exhibited the universal property for $H$ over the full system $(G_{i,j})_{(i,j)\in I\times J}$.
A: Instead of the category of groups you can work in any category. By the Yoneda Lemma it suffices to prove the statement in the category of sets. But there it is easily checked, since all three sets are the same subsets of $\prod_{i,j} G_{i,j}$. You have compatibility in $(i,j)$, iff you have compatibility in $i$ and compatibility in $j$: $\Rightarrow$ follows by considering identities in one entry, and $\Leftarrow$ follows by composition.
