A simplicial set is called finite if it has only a finite number of non degenerate simplicies (or equivalently, if its number of $n$-simplicies grows polynomially with $n$). In an answer to a previous question of mine
[Exponentiation in finite simplicial sets](https://mathoverflow.net/questions/149149/exponentiation-in-finite-simplicial-sets)
it was shown that for any $n\geq 4$ the simplicial set $(\Delta\_n/\partial \Delta\_n)^{\Delta\_1}$ is not finite. Following one of the comments there, I wanted to ask:



> 
> Is the simplicial set $(\Delta\_3/\partial \Delta\_3)^{\Delta\_1}$ finite?
> 
> 
> 



Is the simplicial set $(\Delta_3/\partial \Delta_3)^{\Delta_1}$ finite?
No, it is not; here is a slightly simpler argument than the one indicated in my comment. As in the answer to the previous question, let $X=\Delta\_3/\partial\Delta\_3$. It suffices to show that the cardinality of $A\_n=\operatorname{Hom}(\Delta\_1\times \Delta\_n, X)$ grows faster than any polynomial as a function of $n$. As in that answer, an element of $A\_n$ can be identified with a sequence $(x\_0,\dots,x\_n)$ of elements of $X\_{n+1}$ such that $d\_{i+1}x\_i=d\_{i+1}x\_{i+1}$ for $0\leq i<n$.


Let $B\_n\subset A\_n$ be the set of such sequences for which $d\_{i+1}x\_i=[\partial\Delta\_3]$ is the basepoint of $X\_n$ for all $0\leq i<n$ and $x\_0=x\_n=[\partial\Delta\_3]$ is the basepoint of $X\_{n+1}$. If $(x\_0,\dots,x\_n)\in B\_n$, then for each $0<i<n$, $x\_i$ must be either $[\partial\Delta\_3]$ or the (image in $X$ of) the unique $(n+1)$-simplex $\sigma\_i$ in $\Delta\_3$ such that $\sigma\_i$ is not in $\partial\Delta\_3$ but $d\_i\sigma\_i$ and $d\_{i+1}\sigma\_i$ are in $\partial\Delta\_3$. Explicitly, this simplex is the map $\sigma\_i:[n+1]\to[3]$ such that $\sigma\_i(j)=0$ for $j<i$, $\sigma\_i(i)=1$, $\sigma\_i(i+1)=2$, and $\sigma\_i(j)=3$ for $j>i+1$.


So to define an element of $B\_n$, we have exactly two choices for each $x\_i$ for $0<i<n$, and every possible sequence of such choices does define an element of $B\_n$. Thus $B\_n$ has exactly $2^{n-1}$ elements. In particular, $|A\_n|\geq 2^{n-1}$ grows exponentially.


