Let $X$ and $S$ be simplicial sets, i.e. presheaves on $\Delta$, the so-called topologist's simplex category, which is the category of finite nonempty ordinals with morphisms given by order preserving maps.

How can we derive the structure of the face and degeneracy maps of the join from either of the two equivalent formulas for it below:

The Day Convolution, which extends the monoidal product to the presheaf category:

$$(X\star S)_{n}:=\int^{[c],[c^\prime] \in \Delta_a}X_{c}\times S_{c^\prime}\times Hom_{\Delta_a}([n],[c]\boxplus[c^\prime])$$

Where $\Delta_a$ is the augmented simplex category, and $\boxplus$ denotes the ordinal sum. The augmented simplex category is the category of all finite ordinals (note that this includes the empty ordinal, written $[-1]:=\emptyset$.

The join formula (for $J$ a finite nonempty linearly ordered set):

$$(X\star S)(J)=\coprod_{I\cup I=J}X(I) \times S(I')$$ Where $\forall (i \in I \text{ and } i' \in I'),$ $i < i'$, which implies that $I$ and $I'$ are disjoint.

Then we would like to derive the following formulas for the face maps (and implicitly the degeneracy maps):

The $i$-th face map $d_i : (S\star T)_n \to (S\star T)_{n-1}$ is defined on $S_n$ and $T_n$ using the $i$-th face map on $S$ and $T$. Given $\sigma \in S_j\text{ and }\tau\in T_k$ , we have:

$$d_i (\sigma, \tau) = (d_i \sigma,\tau)\text{ if } i \leq j, j ≠ 0.$$ $$d_i (\sigma, \tau) = (\sigma,d_{i-j-1} \tau) \text{ if } i > j, k ≠ 0.$$ $$d_0(\sigma, \tau) = \tau \in T_{n-1} \subseteq (S\star T)_{n-1} \text{ if } j = 0$$ $$d_n(\sigma, \tau) = \sigma \in S_{n-1} \subset (S\star T)_{n-1}\text{ if } k = 0$$

We note that the special cases at the bottom come directly from the inclusion of augmentation in the formula for the join.

Edit: Another note here: I got these formulas from a different source, so the indexing may be off by a factor of -1.


How can we derive the concrete formulas for the face and degeneracy maps from the definition of the join (I don't want a geometric explanation. There should be a precise algebraic or combinatorial reason why this is the case.)?

Less importantly, how can we show that the two definitions of the join are in fact equivalent?


Ideally, an answer would show how to induce one of the maps by a universal property.

Note also that in the second formula, we allow $I$ or $I'$ to be empty, and we extend the definition of a simplicial set to an augmented simplicial set such that $X([-1])=*$, i.e. the set with one element.

A further note about the first formula for the join: $\boxplus$ denotes the ordinal sum. That is, $[n]\boxplus [m]\cong [n+m+1]$. However, it is important to notice that there is no natural isomorphism $[n]\boxplus [m]\to [m]\boxplus [n]$. That is, there is no way to construct this morphism in a way that is natural in both coordinates of the bifunctor. This is important to note, because without it, it's not clear that the ordinal sum is asymmetrical.

  • $\begingroup$ I am somewhat stymied by the main question. I would not want to give an answer that consists of, e.g., writing down the actual definition of d_i, in terms of being the functor applied to a certain map, and then observing that when [n] is decomposed into two halves there are two possibilities for how d_i might occur. Where are you finding your definition for d_i and can you be more specific about why applying this to the definition of the join is problematic? $\endgroup$ Mar 26, 2010 at 2:46
  • $\begingroup$ Let me put it this way, I have looked through the literature extensively and have not been able to find a real proof over the course of five months, since I asked the first of this series of questions. If you can explain it with an informal combinatorial argument rather than an informal geometric one, it might be sufficient. A geometric argument is not sufficient because it does not connect the symbolic representation above with the geometry. $\endgroup$ Mar 26, 2010 at 3:30
  • 1
    $\begingroup$ My problem with the above definitions of those maps is that there's no obvious way to derive them from the definition. The whole point of the join being a monoidal product is that we should be able to derive the structure maps formally from the definition of the underlying product and the structure maps of the two presheaves. $\endgroup$ Mar 26, 2010 at 3:37
  • $\begingroup$ All right. I don't have time now, but I'll try to write something down tomorrow if you don't have a satisfactory answer. $\endgroup$ Mar 26, 2010 at 4:25
  • $\begingroup$ Thanks! After waiting this long, another day won't hurt. Let's just hope I don't get hit by a bus! =) $\endgroup$ Mar 26, 2010 at 4:33

1 Answer 1


(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.

  • $\begingroup$ Thank you so much, I really appreciate it. This is an awesome answer. $\endgroup$ Mar 26, 2010 at 13:45

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