In the category of simplicial sets there is an 'interchange' map $$h : (A \times B) * (C \times D) \to (A * C) \times (B * D)$$ given by $$h = (\pi_A * \pi_C, \pi_B * \pi_D)$$ where $\times$ is the cartesian product and $*$ is the join. Is it a monomorphism?
I think that it is but I'm not fully confident in my reasoning: if $f,g : Q \to (A \times B) * (C \times D)$ and $x \in Q_n$ then since $$(X * Y)_n = \coprod_{i+j+1=n} X_i \times Y_j$$ we have that $f(x)$ is a '4-tuple' $$f(x) := (f(x)_i, f(x)_j) := ((f(x)_i^a,f(x)_i^b), (f(x)_j^c,f(x)_j^d)) \in (A \times B)_i \times (C \times D)_j$$ for some $i,j$ with $i + j + 1 = n$. Now $$h(f(x)) = ((f(x)_i^a,f(x)_j^c),(f(x)_i^b), f(x)_j^d)) \in (A * C)_n \times (B * D)_n$$ So if $h(f(x)) = h(g(x))$ then $f(x) = g(x)$ since each dimension and component of the 4-tuples $h(f(x))$ and $h(g(x))$ and therefore of $f(x)$ and $g(x)$ must match, so $h$ is a pointwise injection, and therefore a monomorphism?
I'll reproduce my answer from over on Zulip:
Simplicial sets form a presheaf category. In particular, for a map to be a mono, you only need to check that it's componentwise injective. To demonstrate that this is indeed the case for the map you're considering, I'll employ the shortcut of taking the join of augmenting and then truncating the simplicial sets, which simplifies the notation by allowing the indices $i$ and $j$ in what follows to be $\geq -1$.
Explicitly, $$((A \times B) * (C \times D))_n = \coprod_{i+j = n-1} (A_i \times B_i) \times (C_j \times D_j).$$ Meanwhile, $$((A * C) \times (B * D))_n = \left(\coprod_{i+j = n-1} (A_i \times C_j)\right) \times \left(\coprod_{i+j = n-1} (B_i \times C_j) \right).$$ If I'm not mistaken, the map $h$ sends $((a,b),(c,d))$ to $((a,c),(b,d))$, with no identification. This is "obviously" injective, since if two simplices in the domain have equal images, their component simplices must be the same.