Let $X,Y:\mathbb{N}^2\to Ab$ be two functors. Here $\mathbb{N}^2$ is the discrete category on that set, and Ab is the category of abelian groups.
Let $\sigma:X\to Y$ be a natural transformation. Suppose I know that for every $n$, we have that $\sigma$ induces isomorphisms
$ colim_p X(n,p)\to colim_p Y(n,p)$ between "vertical colimits".
Is it true that the "diagonal colimit" map $colim_p X(p,p)\to colim_p Y(p,p)$ is an isomorphism as well?
Let $\phi : \text{colim}_p X(n,p) \to \text{colim}_pY(n,p)$ be the isomorphism given and let $i_p : X(n,p) \to \text{colim}_p X(n,p)$ and $j_p: Y(n,p) \to \text{colim}_p Y(n,p)$ be the $p$th component of the respective cones. Then, by assumption $j_p \sigma_{n,p} = \phi i_p$ and hence $\sigma_{n,p}$ is a monomorphism.
Next, let $u_p : \text{colim}_p X(n,p) \to X(n,p)$ and $v_p : \text{colim}_p Y(n,p) \to Y(n,p)$ be the unique morphisms such that $u_pi_q$ is $0$ when $p\neq q$ and identity when $p=q$, and $v_pj_q$ is $0$ when $p\neq q$ and identity when $p=q$. An easy calculation shows that $\sigma_{n,p} u_p i_q = v_p i_q \sigma_{n,q} = v_p \phi i_q$ and hence by the universal property $\sigma_{n,p} u_p=v_p \phi$. It follows that $\sigma_{n,p}$ is a surjection and hence an isomorphism.
This means that the "diagonal colimit" map is an isomorphism.