I am struggling to understand how to take colimits in the category of simplicial sets. To make my notation clear, I will use $\delta_i : [n-1]\to[n]$ to denote the $i$-th face map (it picks out the face opposite vertex $i$). And I will use $\sigma_i:[n+1]\to [n]$ to denote the $i$-th degeneracy. To illustrate my trouble, here is a specific example that I am struggling with:
I want to show that the following is a pushout square:
First, I'm not even sure what the two unnamed maps are. I think they are $(\delta_2)_*$ on the bottom and $(\delta_0)_*$ on the right. But I'm really not sure.
I know that the category of simplicial sets is a functor category and so colimits can be taken component-wise. It is not too hard to write out $\Delta^0([n])$, $\Delta^1([n])$ and take the colimit of these sets, for small values of $n$. But I have no way to do this for arbitrary values of $n$. E.g., $\Delta^0([1])=\{\sigma_0\}$ and $\Delta^1([1])=\{Id_{[1]},\delta_0\sigma_0,\delta_1\sigma_1\}$. Taking fitting these into the above diagram and taking their pushout lends $\{[Id_{[1]}], [\delta_0\sigma_0]\}$. Here $[Id_{[1]}]$ is the equivalence class $\{Id_{[1]}\}$ and $[\delta_0\sigma_0]$ is the equivalence class $\{\delta_0\sigma_0,\delta_1\sigma_0\}$. Now, since $\Lambda^2_1([1])=\{\delta_2,\delta_0\}$, these two sets are in fact isomorphic (since this is set, this is just a bijection).
But, (1) wouldn't I need to fit this isomorphism into a natural isomorphism so that I can determine the legs of the cocone with $\Lambda^2_1$ at the nadir? I'm not sure I see how this would happen.
And (2) , I'm not sure how this helps me prove that $\Lambda^2_1$ really is the corner of the pushout square. This worked for $[1]$ because things were really easy to write out. Could somebody guide me through this proof, and maybe hint at a general technique for calculating colimits in $Set_\Delta$, component-wise or otherwise?