Smash product of Kan complexes 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.
 A: 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.
