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. 



Joins of simplicial sets
Implicit in the index of the coproduct is that you're writing J as an ordered disjoint union of I and I', where I comes first.


EDIT: Some elaboration.


For a simplicial set $T$, let's write $T\\_n$ for the "n-simplices", i.e. the value of on the ordered set $\{0,1,...,n\}$; these together with the maps between them determine the functor $T$ completely. (Your formula for the join requires the convention that $T$ takes the empty set to a single point.)


Given $S$ and $S'$, let's determine the 0- and 1-simplices of the join.


First, $(S \star S')\\_0$. There are exactly two ways to write $\{0\} = I \cup I'$ in an order preserving way as indexed by the coproduct: either $I'$ is empty and $I$ is everything, or vice versa. Thus $(S \star S')\\_0 = S\\_0 \cup S'\\_0$ accordingly. The zero-simplices of the join are the zero-simplices of the original simplicial set.


Next, the 1-simplicies. Similarly
$$
(S \star S')\\_1 = S(\{0,1\}) \cup (S(\{0\}) \times S'(\{1\})) \cup S'(\{0,1\})= S\\_1 \cup (S\\_0 \times S'\\_0) \cup S'\\_1
$$
There are then 3 types of 1-simplices: the 1-simplices from S, those from S', and for each choice of a point of S and a point of S' there is a new 1-simplex.


The two boundary maps $(S \star S')\\_1 \to (S \star S')\\_0$ are induced by the inclusions of $\{0\}$ and $\{1\}$ into $\{0,1\}$ (the "back" and "front" boundaries respectively). In particular, on the new 1-simplices $S\\_0 \times S'\\_0$ the back boundary is the projection to $S\\_0$ and the front boundary is projection to $S'\\_0$. There is *asymmetry* here because the only ways we're allowed to decompose $\{0,1\}$ in the coproduct have $I$ (the subset corresponding to $S$) first and $I'$ second. None of the "new" edges start at a vertex of $S'$ and end at a vertex of $S$.


