# Simplicial nerve functor commutes with opposites

There are two "opposite" functors: $$op_\Delta\colon sSet\to sSet$$ and $$op_s\colon sCat\to sCat.$$ The first takes a simplicial set to its opposite simplicial set by precomposing with the opposite of a functor $\Delta\to \Delta$ which is the identity on objects and takes a morphism $\langle k_0,\ldots,k_n\rangle\colon [n]\to [m]$ (where $k_i$ is the integer that $i$ gets mapped to by this morphism) to the morphism $\langle m-k_n,\ldots,m-k_0\rangle$. For example, the morphism $[1]\to [2]$ that takes $0$ to $0$ and $1$ to $1$ gets mapped to the morphism that takes $0$ to $1$ and $1$ to $2$.

The second functor takes a simplicial category to the opposite simplicial category, which is easier to define. It has the same objects but given $x,y\in C^{op}$, the mapping complex $C^{op}(x,y)=C(y,x)$.

There is also the simplicial nerve functor $N\colon sCat\to sSet$. I am interested in a proof of the fact that for a given fibrant simplicial category $C$, there is a weak equivalence of quasicategories $op_\Delta\circ N(C)\simeq N\circ op_s(C)$.

I'm relatively certain that this is an elementary proof, but I don't feel skilled enough with the simplicial nerve to figure out the details. Does anyone have a proof of this fact?

## 1 Answer

It all follows from the following elementary lemma:

$\mathfrak{C}([n]^{op})$ is isomorphic to $\mathfrak{C}([n])^{op}$ as a cosimplicial simplicial category (in fact, they are actually equal, since the components of the natural isomorphism are all identities).

proof: It is an immediate calculation from the definition of opposites.

Consequence: From this lemma, apply adjunctions to show that $^{op}$ actually commutes with the simplicial nerve up to isomorphism. I can fill in details if you want.

• For what it's worth, the "immediate calculation" is a little bit involved, depending on your explicit model of $\mathfrak{C}[n]$ and probably requires induction. I have a construction that I haven't seen elsewhere that makes this induction pretty easy, and also the construction in Riehl-Verity of $\mathfrak{C}[\infty]$ should make it easy as well, but the version described in HTT will probably make this calculation very very hard. – Harry Gindi Aug 15 '18 at 4:05
• I'm pretty sure it is enough to check it for $n=0$ and $n=1$ (since $\Delta^0$ and $\Delta^1$ generate $\mathrm{Cat}_\infty$ under colimits), if that helps – Denis Nardin Aug 15 '18 at 8:13
• @DenisNardin It seems like that only works if you use a homotopical approach. For better or for worse, I have answered with a sketch of a combinatorial proof that works at the point-set level. It's not clear to me how knowing that $\Delta[1]$ and $\Delta[0]$ generate $\operatorname{Cat}_\infty$ under (homotopy) colimits helps prove the point-set statement. – Harry Gindi Aug 15 '18 at 8:39
• Well, Jon was asking only whether the two functors are naturally equivalent. For that it is enough to prove that they induce the same equivalence between the two ∞-cats. Of course if you want to prove the point-set statement you need to do more work – Denis Nardin Aug 15 '18 at 9:08
• @DenisNardin I guess the benefit of the point-set statement is that the proof is elementary, the result is stronger, and it doesn't require the dozens of pages needed to prove that $\mathfrak{C}$ and $\mathfrak{N}$ are a QEQ and that QEQs become adjoint equivs under the Relative Nerve. – Harry Gindi Aug 15 '18 at 9:22