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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?

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  • $\begingroup$ I believe that Aaron's comment in the chat room answers this question: chat.stackexchange.com/transcript/message/25359959#25359959 $\endgroup$ – Jonathan Beardsley Nov 12 '15 at 19:38
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    $\begingroup$ If so, then I think the right thing to do is post it as CW, accept it, and don't leave this one unanswered. $\endgroup$ – David White Nov 12 '15 at 20:41
  • $\begingroup$ @DavidWhite I don't even know how to turn a post into a Community Wiki, much less accept a comment as an answer. $\endgroup$ – Jonathan Beardsley Nov 12 '15 at 23:08
  • $\begingroup$ I suppose it'd be better if @Aaron could come answer it. But apparently I can't ping him from here. $\endgroup$ – Jonathan Beardsley Nov 12 '15 at 23:11
  • $\begingroup$ No, I mean you should just copy and paste what he wrote into the answer field below and then accept it (so you answer your own question, with an acknowledgment to Aaron). To make an answer CW just click the box in the corner. $\endgroup$ – David White Nov 12 '15 at 23:22
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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.

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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.

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