It turns out that joins of simplicial sets are fairly easy to define, but hard to manage. In lots of cases, we'd like to compute what a join is, does it look like a horn?, a boundary?, etc? and identify it as such, so we can figure out when our morphisms from the join have certain nice properties like being anodyne, having lifting properties, and all of that wonderful stuff.


For example, consider the join, $\Lambda^n\_j \star \Delta^m$. The problem that I currently face is, I can't tell what this thing looks like from the definition. 


Consider an even simpler case, $\Delta^n \star \partial \Delta^m$. From the definition, we get a very nasty definition of this join, and I'm having trouble applying it and computing the join in terms of nicer simplicial sets. 


I ask this, because on p.62 of Higher Topos Theory by Lurie, for example, he states that for some $0 < j \leq n$ $$\Lambda^n\_j \star \Delta^m \coprod\_{\Lambda^n\_j \star \partial \Delta^m} \Delta^n \star \partial \Delta^m$$
and says that we can identify this with the horn $\Lambda^{n+m+1}\_j$. Unraveling the definitions seems to make it harder to understand, and I just don't see how this result was achieved. However, my aim here is to understand how the computation was actually carried out, since it is completely omitted. 


For convenience, here is the definition of the join of $S$ and $S'$ for each object $J \in \Delta$
$$(S\star S')(J)=\coprod\_{J=I\cup I'}S(I) \times S'(I')$$
Where $\forall (i \in I \land i' \in I') i < i'$, which implies that $I$ and $I'$ are disjoint. 


**EDIT AFTER ANSWER:** Both Reid and Greg provided good answers to the question, and I only accepted the one that I did because Greg commented more recently. So for anyone reading this at some point in the future, read both answers, as they are both good.



Computation of Joins of Simplicial Sets
Since the join of simplicial sets is associative and $\Delta^m = \Delta^0 \star \cdots \star \Delta^0$ ($m+1$ times), we should start by trying to understand things like $\Lambda^n\_j \star \Delta^0$, a.k.a. the "final" cone on $\Lambda^n\_j$. It's not too hard to see that this is the subcomplex of $\Delta^{n+1}$ consisting of those faces which do not contain the (codimension 2) face $\{0, \ldots, r-1, r+1, \ldots, n\}$. In other words, we are missing the face opposite $r$ and $n+1$, because we were originally missing the face opposite $r$ of $\Delta^n$, as well as the three other faces (including the interior of $\Delta^{n+1}$) it contains. Similarly $\Delta^0 \star \partial \Delta^n$ is the horn $\Lambda^{n+1}\_0$ (we are missing the interior of $\Delta^{1,\ldots,n}$ and the cone on it).


In general all the simplicial sets that come up have the form of the subcomplex of $\Delta^N$ consisting of those faces which do not contain a fixed face $\Delta^S$, $S \subset \{0, \ldots, N\}$. Forming the cone (on either side) on such a space results in another such space with $N$ replaced by $N+1$ and $S$ unchanged (as a subset of the vertices of the original $\Delta^N$, which if we formed a cone on the left, means we increment each index in $S$).


After doing these sorts of computations, I expect that $\Lambda^n\_j \star \Delta^m$ and $\Delta^n \star \partial \Delta^m$ will be two subcomplexes of $\Delta^{n+m+1}$ each characterized by avoiding faces containing a certain face, and that $\Lambda^n\_j \star \partial \Delta^m$ is their intersection and $\Lambda^{n+m+1}\_j$ is their union, from which the claim would follow.


