Relation between limit on a grid and limit of the diagonal

Question

Let $\mathcal{C}$ be a category with all limits and suppose we have a functor $$F:\mathbb{N}^2\longrightarrow\mathcal{C}\ ,$$ where $\mathbb{N}^2$ is seen as the category with morphisms generated by $(n+1,m)\to(n,m)$ and $(n,m+1)\to(n,m)$. Can anything be said about the relations between $$\lim_{n,m}F(n,m)\qquad\text{and}\qquad\lim_k F(k,k)?$$ In particular, are there assumptions one can make so that they are naturally isomorphic? I tried drawing a couple diagrams, but I got nowhere.

Answer 1

No assumption is needed. You have a canonical morphism $$\displaystyle\lim_{n,m} F(n,m) \to \lim_{k} F(k,k)$$ constructed as follow: take the canonical cone $(\lim_{n,m} F(n,m) \to F(k,\ell))_{k, \ell}$ and forget about the non-diagonal terms, you end up with a cone $(\lim_{n,m} F(n,m) \to F(k,k))_{k}$ that must factor through the limit $\lim_k F(k,k)$.

The claim is that this canonical map is always iso.

Not very formally, it goes as follow: to define the inverse map is the same as defining a specific cone $(\lim_k F(k,k) \to F(n,m))_{n,m}$ which is constructed at component $(n,m)$ by selecting any $p\geq \max(n,m)$ and composing $\lim_k F(k,k) \to F(p,p)$ with $F(p,p) \to F(n,m)$. You can check that this i s well defined and that it induces indeed an inverse for the canonical map above.

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But you can also call the big guns because this is know in greater generality. The previous claim is equivalent to this one:

The diagonal functor $d:\mathbb N \to \mathbb N^2$ is co-cofinal.

So all we have to check is the following property: for each $(n,m)$, the comma category $(d\downarrow (n,m))$ is non empty and connected. This is the case:

1. for any $(n,m)$, any $p \geq \max(n,m)$ gives an object $(p,p)\to (n,m)$ of the category $(d \downarrow(n,m))$, hence the non-emptyness
2. for two such objects $(p,p)\to (n,m)$ and $(q,q)\to (n,m)$, select the greater between $p$ and $q$ (say $q$ here), giving a morphism $(q,q) \to (p,p)$ of the category $(d \downarrow(n,m))$, hence the connectedness.