(A note: I am going to regard simplicial sets as also defined on the empty ordinal as well, with $X(\emptyset) = *$, which is required for the join formula. This is implicit in your first definition and will remove the need for two extra cases for $d_i$ at the end.)
Regarding the "minor" question. The short explanation is that this follows by decomposing the Hom-set according to the preimage of $c$ and $c'$ in $n$, and observing that each decomposition of $n$ provides an initial choice.
In more category-theoretic language, one way to rewrite the convolution is using the "over" category:
$$
(X \star S)_n = \int^{[n] \to [c] \boxplus [c']} X_c \times S_{c'}
$$
where now the coend is taken over the comma category $n \downarrow \boxplus$ whose objects are triples $([c],[c'],f)$ of a pair of objects of $\Delta$ and a morphism from $[n]$ to their ordinal sum. We note that this comma category decomposes as a disjoint union of categories: each $([c],[c'],f)$ determines a decomposition $[n] = f^{-1} [c] \cup f^{-1} [c']$ into a disjoint union, and morphisms preserve such a decomposition. Therefore,
$$
[n] \downarrow \boxplus \simeq \coprod_{[n] = I \cup I'} (I \downarrow \Delta) \times (I' \downarrow \Delta)
$$
This makes the coend decompose:
$$
(X \star S)_n = \coprod_{[n] = I \cup I'} \int^{I \to [c], I' \to [c']} X_c \times S_{c'}
$$
However, the comma category $(I \downarrow \Delta)\times(I' \downarrow \Delta)$ has an initial object: $I \times I'$ itself. Thus, the coend degenerates down to simply being the value:
$$
(X \star S)_n = \coprod_{[n] = I \cup I'} X_{|I|} \times S_{|I'|}
$$
This is slightly different notation for the second definition of the join that you gave.
Now, as for the boundary formulas.
The definition of $d_i$ is as follows. For each $0 \leq i \leq n$, there is a unique map $d^i:[n-1] \to [n]$ in $\Delta$ whose image is $[n] \setminus \{i\}$: $d^i(x) = x$ for $x < i$, and $d^i(x) = x+1$ for $x \geq i$. The induced map $(X \star S)_n \to (X \star S)_{n-1}$ is the map induced by applying the contravariant functor to $d^i$.
Since $(X \star S)_n$ is a disjoint union of sets, it suffices to show that the formula is correct on $X(I) \times S(I')$ for all decompositions of $[n]$ into $I \cup I'$, where $|I| = j+1$ and $|I'| = k+1$. There are two possibilities: either $i \in I$ when $0 \leq i \leq j$, or $i \in I'$ when $j < i \leq n$.
In either case, the map $[n-1] \to [n] = I \cup I'$ induces, by taking preimages, a unique ordered decomposition $[n-1] = J \cup J'$ of $[n-1]$. If $i \in I$, then $J$ has size $|I| - 1$ and $J'$ is mapped isomorphically to $I'$ by $d^i$. In this case, the map $d^i$ is isomorphic to the map $d^i \boxplus id$ on $[j-1] \boxplus [k] \to [j] \boxplus [k]$. If $i \in I'$, we have the reverse possibility, with $d^i$ isomorphic to $id \boxplus d^{i-j-1}$ (the upper index necessary because inserting the identity at the beginning adds $j+1$ elements to the ordered set at the beginning).
In the case $0 \leq i \leq j$, the induced map
$$
d_i: X(I) \times S(I') \to \coprod_{[n-1] = K \cup K'} X(K) \times S(K')
$$
is therefore the map $X(d^i) \times id: X(I) \times S(I') \to X(J) \times S(J')$, followed by the inclusion into the coproduct. In the case $j < i \leq n$, the map is $id \times S(d^{i-n-1})$ followed by inclusion.
This recovers the formula for $d_i$ that you have written down, up to inserting copies of a point $*$ as in the remark at the beginning.
Marco Grandis has done some work on this, and you can extract answers for 1-3 from these papers
- Finite Sets and Symmetric Simplicial Sets - M Grandis - TAC (pdf)
- Higher Fundamental Functors for Simplicial Sets - M Grandis - CTGDC (pdf)
See also
- An Alternative Presentation of the Symmetric-Simplicial Category - Eric R. Antokoletz - arxiv (link)
Question 1 and 2
The first paper by Grandis above gives a nice detailed overview of all this, including
- a description of (a skeleton of) $\mathbf{FinSet}$ as the walking symmetric strict monoidal category with a commutative monoid
- in terms of generators and relations, with generators faces + degeneracies + transpositions, and relations the standard ones for faces + degeneracies, the Moore ones for transpositions, and some compatibility rels between those.
About who first proved this kind of things, in the second paper, he acknowledges that:
In November 1998, at a PSSL meeting in Trieste, Bill
Lawvere suggested I might extend my study of the homotopy of simplicial
complexes to symmetric simplicial sets, on the basis of his draft [19] where the
fundamental groupoid of the latter is presented as a left adjoint. I would like to
express my gratitude for his kind encouragement and for helpful discussions.
[19] is Lawvere's Toposes generated by codiscrete objects, in combinatorial topology and functional analysis from 1989, which I don't have access to; maybe there's some more info in there.
Question 3
You could find more about this in the second paper by Grandis. The idea is that the classical homotopy theory of simplicial complexes can be extended to symmetric simplicial sets (presheaves on $\mathbf{FinSet}$), so that the edge-path groupoid of a simplicial complex can be identified with what Grandis calls the intrinsic fundamental groupoid, which is the left adjoint of a symmetric nerve (this goes back to Lawvere notes ref above). See also
- An intrinsic homotopy theory for simplicial complexes, with applications to image analysis - M Grandis (pdf)
Best Answer
Here is half of a classification. Let $\otimes$ be a monoidal structure on $\Delta_+$. As I mentioned in a comment, the monoidal unit must be $[-1]$ or $[0]$ because these are the only objects with commutative endomorphism monoids.
Suppose that the monoidal unit is $[0]$. Let us consider $[1] \otimes [1]$. We have that $[0]$ is a retract of $[1]$ in 2 ways, and as a result we obtain 4 retracts of $[1] \otimes [1]$ with support $[0] \otimes [0] = [0]$. Consider the induced linear ordering on these 4 points. We also have 4 ways that $[1] = [1] \otimes [0] = [0] \otimes [1]$ is a retract of $[1] \otimes [1]$, and from this we can deduce most of the ordering. It must have the following relations
$\require{AMScd} \begin{CD} 0 \otimes 0 @>>> 0 \otimes 1\\ @VVV @VVV\\ 1 \otimes 0 @>>> 1 \otimes 1 \end{CD}$
To complete this to a linear order, without loss of generality we must have $0 \otimes 1 \leq 1 \otimes 0$. But now, one of our 4 projections onto $[1]$ is the coordinate projection onto the right column of the above square. The fact that this projection is order-preserving implies that $1 \otimes 0 = 0 \otimes 1 = 1 \otimes 1$. This contradicts the fact that the right column exhibits $[1]$ as a retract of this subset of $[1] \otimes [1]$.
Therefore the monoidal unit is not $[0]$; it must be $[-1]$.
I also think I'm ready to conjecture that $\oplus$, $\oplus^{rev}$ (as mentioned by Peter) and the degenerate monoidal structure are probably the only ones. You could imagine a classification starting as follows. Consider the maps $[0] = [0] \otimes [-1] \to [0] \otimes [0]$ and $[0] = [-1] \otimes [0] \to [0] \otimes [0]$. If these are the same, then we should have the degenerate monoidal structure. If they are different, then one is less than the other, and those two cases should correspond to $\oplus$ and $\oplus^{rev}$.