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?



Exponentiation in finite simplicial sets
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.


