Suppose $X$ and $Y$ are pointed Kan complexes. Is their smash product $X\wedge Y$ also a Kan complex?


I expect the answer is probably no, but it would be nice to see a counterexample.



Smash product of Kan complexes
The basic idea is the map $X \times Y \to X \wedge Y$ is a bijection on simplices except over the basepoint. We can construct an extension problem that doesn't lift to an extension problem on the cartesian product.


Let $X$ and $Y$ both be the standard model for $E\mathbb{Z}/2$, which is the nerve of a groupoid with two isomorphic objects and trivial automorphism groups. Call the basepoint $\ast$ and the nonbasepoint $p$; call $a$ the edge from $\ast$ to $p$ and $b$ its inverse. In simplicial identities, $d\_1(a) = \ast, d\_0(a) = p$, and the reverse for $b$.


We can construct a map from the inner $2$-horn $\Lambda\_1^2$ of $X \wedge Y$ with no horn filler as follows. The $1$-simplex $b \wedge a$ satisfies
$$
d\_0(b \wedge a) = \ast \wedge p = \ast = p \wedge \ast = d\_1(b \wedge a),
$$
and so it defines a horn.


However, there is no horn filler. If there was, we would have a $2$-simplex of $X \wedge Y$ filling it. As mentioned, the map $X \times Y \to X \wedge Y$ is surjective on $2$-simplices, and injective except on those simplices living over the basepoint. Therefore, we would have a $2$-simplex of $X \times Y$ satisfying $d\_0(u) = d\_2(u) = b \times a$. However, in $X \times Y$ we have
$$
d\_0(b \times a) = \ast \times p \neq p \times \ast = d\_1(b \times a),
$$
and so there can be no such 2-simplex.


