Kerodon

The last map should read $\circ[\beta]$ instead of $\circ[\alpha]$.

Yep. Thanks!

I have a question regarding the following specific colimit diagram: Let $K$ be the simplicial set $\Delta^1 \coprod_{\Delta^0} \Delta^1$, which is isomorphic to the nerve associated to the preorder containing three objects $0,1,2$ with $0<1$ and $0<2$. Let $u: K \rightarrow \mathcal{S}$ be the diagram given by $u(0)=J$, $u(1)=u(2)=\Delta^0$, here $\mathcal{S}$ is the $\infty$-category of Kan-complexes and $J$ is the interval obtained by inverting the non-trivial morphism in $\Delta^1$. We know that $J$ is weak homotopy equivalent to $Sing(S^1)$ and the (homotopy) colimit of the diagram $u$ should be isomorphic to $Sing(S^2)$. In the following I write $S^1$ and $S^2$ to mean $Sing(S^1)$ and $Sing(S^2)$. Now by the definition of colimit above, we should have an isomorphim of Kan-complexes $Hom_{\mathcal{S}}(S^2,S^2) \rightarrow Hom_{\operatorname{Fun}(K,\mathcal{S})}(u,\underline{S}^2)$. The left hand side has $\pi_0$ isomorphic to $\mathbb{Z}$. The right hand side is a subcomplex of $\operatorname{Fun}(K \times \Delta^1,\mathcal{S})$ with vertices $f:K \times \Delta^1 \rightarrow \mathcal{S}$ which restricts to $u$ and $\underline{S}^2$ on the two ends. But it seem to me that there is only one such morphism $f$, thus the right hand side would have trivial $\pi_0$(I think I was wrong in computing this morphism space). So I want know how to understand this example correctly.

the misplaced part above is "0,1,2 with 0<1 and 0<2"