TITLE: Joins of simplicial sets
QUESTION [7 upvotes]: Why doesn't the join operation on the category of simplicial sets commute up to unique isomorphism?  I mean, aren't products and coproducts commutative up to isomorphism? That leads me to conclude at first glance that the join is commutative, but it's not. Recall, given two simplicial sets $S$ and $S'$, we define the join to be the simplicial set such that for all finite nonempty totally ordered sets $J$,  $$(S\star S')(J)=\coprod_{J=I\cup I'}S(I) \times S'(I')$$
Where $\forall (i \in I \land i' \in I') i < i'$, which implies that $I$ and $I'$ are disjoint.  
Now the thing is, clearly my conclusion is stupid, because we use the fact that it doesn't commute to distinguish between over quasi-categories and under quasi-categories.  Where did I go wrong?
I hope this is up to the standards of MO, but if it's not, I'll delete the topic.

REPLY [5 votes]: As mentioned in the other answers, the join of simplicial sets is closely related to "ordered disjoint union", or "concatenation", of (totally, partially, pre-) ordered sets.  You can use this both to get simple examples of its non-commutativity, and to help reconcile that with the intuition that it should be commutative.
Any order $X$ can be seen as the simplicial set whose $n$-simplices are chains $(x_0 \leq \ldots \leq x_n)$ from $X$.  That is, there's a full and faithful "nerve" embedding $N: \mathrm{PreOrd} \rightarrow \mathrm{SSet}$.  Now if $X$ and $Y$ are  orders, seen as their nerves, $X \star Y$ is exactly (the nreve of) their ordered disjoint union.
So e.g. $1 \star \mathbb{N} \not \cong \mathbb{N} \star 1$ is an easy and intuitive example of the non-commutativity.
On the other hand, like you say, looking at the definition, there is an immediate intuition that it should be commutative in some sense, and chasing it down, I think what that intuition is coming from is something like the fact: for any simplicial sets $X$, $Y$,
$(X \star Y)^\mathrm{op} \cong Y^\mathrm{op} \star X^\mathrm{op}.$
So commuting $\star$ distributes over $\mathrm{op}$: so in a sense, the only asymmetry in $\star$ is an asymmetry of variance.  This is nice and intuitive for ordered sets, and easily shown for all simplicial sets.