TITLE: Exponentiation in finite simplicial sets
QUESTION [12 upvotes]: 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?

REPLY [7 votes]: 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 non-degenerate 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.