$2$-skeleton of k-th horn of $\Delta^n$

Question

I'm trying to show that $Sk_2(\Lambda_k^n)=Sk_2(\partial \Delta^n)=Sk_2(\Delta^n)$ for $n \geq 4$

is it true for $n=3$? any solution or reference is very much appreciated.

Answer 1

A p-simplex in $\Delta^n$ is represented by a map $[p]\rightarrow [n]$ in the simplex category $\mathbf{\Delta}$. Now observe that for $n\geq 3$ and $p\leq 2$, any map $[p]\rightarrow[n]$ factors through some coface maps $\delta_i:[n-1]\rightarrow [n]$. Since the boundary $\partial\Delta^n$ is the simplicial subset generated by the images of $(\delta_i)_*:\Delta^{n-1}\rightarrow\Delta^n$ we conclude that for such $n$, $p$

$$(\partial\Delta^n)_p=(\Lambda^n_k)_p$$

Hence the two simplicial sets $\Delta^n$ and $\partial\Delta^n$ have the same $0,1,2$ simplices, so that for $n\geq 3$ we have

$$sk_2(\partial\Delta^n)=sk_2(\Delta^n).$$

Now observe that if $n\geq 4$ and $p\leq 2$, then any map $[2]\rightarrow [n]$ in $\mathbf{\Delta}$ factors through at least two different coface maps $\delta_i:[n-1]\rightarrow [n]$. In particular, whatever value of $k$ we choose, we can always assume that $i\neq k$. Hence for $n\geq 4$ and $p\leq 2$

$$(\Lambda^n_k)_p\cong (\Delta^n)_p$$

so again the two simplicial sets $\Lambda^n_k$ and $\Delta^n$ have the same $0,1,2$ simplices. Hence for $n\geq 4$ we have

$$sk_2(\Lambda^n_k)= sk_2(\Delta^n).$$

Finally, as mentioned in the comments, whilst $sk_2(\partial\Delta^n)=sk_2(\Delta^n)$ holds for $n\geq 3$, the equality $sk_2(\Lambda^n_k)= sk_2(\Delta^n)$ holds only for $n\geq 4$. You should be able to draw pictures of the $n=3$ case to convince yourself that it fails.