Definition of Left Operadic Kan Extension for $\infty$-operads

Question

In Lurie's book Higher Algebra, he makes the following definition:

Definition 3.1.2.2: Let $M^\otimes\to N(Fin_\ast)\times\Delta^1$ be a correspondence from an $\infty$-operad $A^\otimes$ to another $\infty$-operad $B^\otimes$, let $q:C^\otimes\to O^\otimes$ be a fibration of $\infty$-operads and let $\overline{F}:M^\otimes\to C^\otimes$ be a map of generalized $\infty$-operads. We say that $\overline{F}$ is an operadic $q$-left Kan extension of $F\vert A^\otimes$ if the following condition is satisfied for every $b\in B^\otimes$:

($\ast$) Let $K=(M_{act}^\otimes)_{/b}\times_{M^\otimes} A^\otimes.$ Then the composite map $$K^\vartriangleright\to (M^\otimes)_{/b}^\vartriangleright\to M^\otimes\overset{\overline{F}}\to C^\otimes$$ is an operadic $q$-colimit diagram.

My question is the following:

What is the map $(M^\otimes)_{/b}^\vartriangleright\to M^\otimes$? Does the cone point of the left hand side have a representative inside of $M^\otimes$? Does the existence of such a map implicitly require such a point to be in $M^\otimes$ already?

Answer 1

Aaron's chat room remark is right: when restricted to $(M^\otimes)_{/b}$ it is the canonical projection, and the cone point is sent to $b$.

Notice that for each object in the slice $f : x \to b$, there is a unique morphism from $f$ to the cone point in $(M^\otimes)_{/b}^\vartriangleright$; this unique morphism is sent to $f$.

To describe the functor fully explicitly as a map of simplicial sets:

If an $n$-simplex $\sigma$ in $(M^\otimes)_{/b}^\vartriangleright$ uses the cone point, the cone point comes at the end, so if the vertices of $\sigma$ are $v_0, \ldots, v_n$, there is some $k \le n$ such that the $v_i$ with $i>k$ are the cone point and the $v_i$ with $i \le k$ are not. Then the face on vertices $\{0, \ldots, k\}$ is a $k$-simplex in $(M^\otimes)_{/b}$, and thus really a $(k+1)$-simplex $\tau$ in $M^\otimes$ with last vertex $b$. The required map is defined in two cases:

• if $k=n$, it sends $\sigma$ to $d_{n+1}\tau$, the face of $\tau$ on $\{0, \ldots, n\}$, and
• if $k<n$, it sends $\sigma$ to $s_{n-1} \cdots s_{k+2} s_{k+1} \tau$, the degenerate simplex on $\tau$ obtained by applying the last possible degeneracy $n-k-1$ times.

Also, the definition of this map works for $M^\otimes$ an arbitrary quasicategory and doesn't require it to be a correspondence between operads.

Answer 2

A recommendation: make drawings of cones! Note that by definition

$$ (X_{/b})_n = Hom_b((\Delta^n)^{\triangleright},X)$$

Where $Hom_b$ means maps that sends the cone point to b. By Yoneda Lemma we can rewrite it as

$$ Hom(\Delta^n, X_{/b}) = Hom_b((\Delta^n)^{\triangleright},X)$$

Now this equation is cocontinous in $\Delta^n$, so it extends to every sSet K, yielding an adjunction:

$$ Hom(K, X_{/b}) = Hom_b(K^{\triangleright},X)$$

In particular, for $K=X_{/b}$ the identity on the left gives a counit $(X_{/b})^{\triangleright} \to X$. Note that you can trace back all the passages to get an explicit description: in the adjunction, you know that every n-simplex $\Delta^n \to K \to X_{/b}$ corresponds to a n+1 simplex $(\Delta^n)^{\triangleright}\to X$ by the second equation.