Inverse limit of a bi-inverse system and limit of the "diagonal" inverse system. I've a question concerning inverse limits, since I don't usually work with them this extensively.
I'm considering the inverse limit of the following "bi-inverse system" of $R$-modules and black arrows $f_{\bullet,\bullet}$, and $g_{\bullet,\bullet}$. Since inverse limits commutes among themselves we can define $$A := \varprojlim_{i,j}A_{i,j}$$ regardless of the order we take them. Let $$\widetilde{A}:=\varprojlim_k A_{k,k}$$ be the limit of the diagonal of this inverse system.

It's easy to see we have a unique map $$\widetilde{A} \to A$$ that makes everything commute (induced by the obvious maps in the system).

I was wondering if it's "easy to see" if there's a map going in the other direction and then prove that $A\cong \widetilde{A}$ maybe.

 A: A functor $j\colon C\to D$ between small categories is initial if pulling back limits along it does not alter limits (in the precise sense that the evident induced morphism is an isomorphism). So, you are asking whether the diagonal inclusion in your case is initial.
If we write $N$ for the category whose objects are the naturals, and with $x\to y$ precisely when $x\ge y$, then your diagram inclusion is the diagonal $\Delta \colon N\to N\times N$. So, the question is whether $\Delta$ is initial. This is the case, and can be verified either directly from the definition: $\Delta$ is initial if for any object $(m,n)\in N\times N$ the slice category $\Delta/(m,n)$ is connected. Alternatively, every left adjoint functor is initial, and $\Delta$ is a left adjoint, its right adjoint is given by the join function $(m,n)\mapsto m\vee n$ (the max function).
A: Just for the sake of argument, it is indeed easy to give a more concrete approach here. The limit of your double system $A$ is a submodule $\prod_{i,j} A_{i,j}$ defined by those tuples $(a_{i,j})$ such that $a_{i-1,j}=f_{i,j}(a_{i,j})$ and similarly for the $g$'s. The canonical map $A\to \widetilde A$ simply sends $(a_{i,j})$ to $(a_{k,k}).$
