A finite simplicial set is a simplicial set having only a finite number of non degenerate simplicies. My question is: if $A$ and $B$ are finite simplicial sets, does this imply that the simplicial set $A^B$ is also finite?

4$\begingroup$ Regarding the tag 'homotopytheory', this question does not appear to be actually about homotopy theory. $\endgroup$ – Ricardo Andrade Nov 17 '13 at 22:27

1$\begingroup$ Maybe combinatorics. $\endgroup$ – Fernando Muro Mar 4 '15 at 23:42
I think the statement is false for $ X = A = \Delta_4/\partial \Delta_4$ and $B = \Delta_1$:
As Charles Rezk already mentioned, it is enough to consider the growth of $f_{X^{\Delta_1}}(n) = Hom(\Delta_1\times \Delta_n, X)$. Writing the prism $\Delta_1 \times \Delta_n$ as a coequalizer over its $(n+1)$ simplices, I think we can identify $Hom(\Delta_1\times \Delta_n, X)$ with tuples $(x_0, \dotsc, x_n)$ where $x_i \in X_{n+1}$ and $ d_{i+1} x_i = d_{i+1} x_{i+1}$ whenever this makes sense. Now, we can restrict our attention to tuples where even $d_{i+1} x_i = d_{i} x_i = [\partial \Delta_4]$ holds (and for simplicity $x_n = [\partial \Delta_4]$ where $[\partial \Delta_4]$ denotes the unique degeneracy of the basepoint).
These definitely satisfy the gluing conditions, and the number of them grows exponentially: We can choose the $x_i$ independently, and for $n$ large enough there are at least 2 possibly choices for any $i< n$: If we denote the unique nondegenerate simplex in $X$ by $\iota$, then we can consider morphisms in the $\Delta$category $\sigma : [n] \rightarrow [4]$ which are surjective and have the property that $\sigma^{1}(\sigma(i))$ and $\sigma^{1}(\sigma(i+1))$ are singleton sets. Then $\sigma^* \iota$ is a possible choice for $x_i$.
This gives exponential growth as a lower bound, so $X^{\Delta_1}$ can't be finite.

$\begingroup$ Very nice! I guess in the argument I had in mind on my comment on Charles's answer I was implicitly imagining $X$ was actually a simplicial complex. $\endgroup$ – Eric Wofsey Mar 4 '15 at 23:24

$\begingroup$ I think you can also get exponential growth for $X=\Delta_3/\partial \Delta_3$ by a slightly different construction. Don't require $d_{i+1}x_i=d_{i+1}x_{i+1}$ to always be $[\partial \Delta_3]$, and instead consider the set $S$ of all values of $i$ such that it is $[\partial\Delta_3]$. It is not hard to see that $S$ can be almost any subset of $\{0,\dots,n\}$, so $Hom(\Delta_1\times \Delta_n, X)$ must have at least $\approx 2^n$ elements. $\endgroup$ – Eric Wofsey Mar 4 '15 at 23:45

$\begingroup$ Thanks! I think I understand the argument, I just don't see why you had to assume $x_n = [\partial \Delta_4]$. I also believe it should be $\sigma : [n+1] \rightarrow [4]$ (or perhaps I misunderstand something?). $\endgroup$ – Ilan Barnea Jul 14 '15 at 6:16
I failed to figure this out. But here are some thoughts.
Say a simplicial set $X$ is finite type if it has finitely many simplices in each degree. It's not hard to prove that $X^K$ is finite type if $X$ is finite type and $K$ is finite.
For finite type $X$, let $f=f_X$ be the function $f_X(n)=X_n$, the number of $n$simplices, and let $g=g_X(n)$ be the number of nondegenerate $n$simplices. Note that $X$ is finite iff $g(n)=0$ for $n$ large.
There is a simple relationship between $f$ and $g$: $$ f(n) = \sum_{k=0}^n \binom{n}{k}g(k).$$ Since the $g(k)\geq0$, we should thus be able to say that $X$ if finite if and only if $f$ has polynomial growth, since $n\mapsto \binom{n}{k}=\frac{1}{k!}n^k+\cdots$ is polynomial of degree $k$ (and is a monotone function).
If $K$ is finite, then $K$ is a union of a finite number of nondegenerate simplices. Thus $X^K\subseteq \prod_{i=1}^d X^{\Delta_{k_i}}$. Since a finite product of finite simplicial sets is finite, it suffices to show that $X$ finite implies $X^{\Delta_k}$ finite.
It's an easy exercise to show that $\Delta_k$ is a retract of the cube $(\Delta_1)^k$. Thus, by the exponential property of function complexes, it suffices merely to show that $X$ finite implies $X^{\Delta_1}$ finite.
A naive estimate is as follows. $f_{X^{\Delta_1}}(n) = Hom(\Delta_1\times \Delta_n, X)$. The complex $\Delta_1\times \Delta_n$ is a union of $n+1$ copies of $\Delta_{n+1}$ along various faces. Thus there is an injection $$ Hom(\Delta_1\times \Delta_n ,X) \subseteq (X_{n+1})^{n+1}.$$ This provides an estimate $$ f_{X^{\Delta_1}}(n) \leq f_X(n+1)^{n+1}. $$ Unfortunately, the right hand side grows exponentially as a function of $n$, so this is of no use.
Note that in some cases it is easy to show that $X^K$ is finite. For instance, if $X$ is the nerve of a finite poset, then $X^{\Delta_n}$ is also the nerve of a finite poset, from which it follows that $X^K$ is finite for all finite $K$. This applies for instance to $X=\Delta_k$.

$\begingroup$ I don't have the energy now to write out the messy details, but I believe that your "naive estimate" can be improved to $O(n^{2dim(X)})$. For large $n$, the $(n+1)$simplices that make up a prism $\Delta_1\times \Delta_n\to X$ must all be very degenerate and very similar to each other; once you pick the first one (about $n^{dim(X)}$ choices), the other ones must more or less look the same except that the locations of the degeneracies get to move around (which I think gives another $n^{dim(X)}$ or so choices). In particular, $X^{\Delta_1}$ should have dimension at most twice that of $X$. $\endgroup$ – Eric Wofsey Nov 17 '13 at 17:55