In the introduction to Chapter 2 of Jacob Lurie's Higher Algebra (entitled "$\infty$-Operads"), a category $\mathcal{C}^\otimes$ is constructed from an arbitrary symmetric monoidal category $\mathcal{C}$. Notably the objects of $\mathcal{C}^\otimes$ are finite lists in the objects of $\mathcal{C}$, and so there is an evident forgetful functor $p : \mathcal{C} \to \textsf{Fin}_*$ with codomain category the pointed finite sets realised as the sets $\langle n \rangle = \{0, 1, \ldots, n\}$ with $n$ always the point.

Lurie gives two important properties of this construction, and I have been stumped at a very superficial level by the second (called "M2"). My confusion is the following: I understand that the fibre $\mathcal{C}^\otimes_{\langle 1 \rangle}$ of $p$ over $\langle 1 \rangle$ is an ordinary category, but fail to see how $\mathcal{C}^\otimes_{\langle 1 \rangle}$ is equivalent to $\mathcal{C}$. In particular, I see a faithful embedding $$ F : \mathcal{C} \to \mathcal{C}^\otimes_{\langle 1 \rangle} $$ defined on objects by $F(C) = [C]$ (the list containing just $C$) and on morphisms by sending $f : C \to D$ to the identity map $\alpha : \langle 1 \rangle \to \langle 1 \rangle$ along with $f_1 = f : C \to D$ (this follows Lurie's notation for the definition of the morphisms of the category $\mathcal{C}^\otimes$).

However, in the situation as I have described it the functor $F$ cannot possibly be full, since we miss all of the morphisms attached to the map of finite pointed sets $\langle 1 \rangle \to \langle 1 \rangle$ defined by $n \mapsto 0$. Does Lurie really mean "equivalence" of categories in the ordinary sense? Have I just completely misunderstood the situation?


If $F : \mathcal{C} \to \mathcal{D}$ is a functor, then the fiber over $d \in \mathcal{D}$ is the subcategory $\mathcal{C}_d$ of $\mathcal{C}$ consisting of objects $c \in \mathcal{C}$ such that $F(c) = d$, and of morphisms $f : c \to c'$ such that the image of $f$ under $F$ is $\operatorname{id}_d$. In your case, $\mathcal{C}^\otimes_{\langle1\rangle}$ is the subcategory of $\mathcal{C}^\otimes$ whose objects are lists with one elements and whose morphisms cover the identity of $\langle1\rangle$. So the embedding $\mathcal{C} \to \mathcal{C}^\otimes_{\langle1\rangle}$ is indeed full (and faithful, and essentially surjective).

I don't know where it is defined in Higher Algebra, but anyway, here's a justification. As all fibers, the fiber $\mathcal{C}_d$ fits in a cartesian square: $$\require{AMScd} \begin{CD} \mathcal{C}_d @>>> \mathcal{C} \\ @VVV @VV{F}V \\ 1 @>d>> \mathcal{D} \end{CD}$$ where $1$ is the terminal category, with one object and one (identity) morphism, and $1 \to \mathcal{C}$ maps the unique object to $d$. If you work it out, you see that not only do the objects of $\mathcal{C}_d$ have to map to $d$, but also the morphisms of $\mathcal{C}_d$ have to map to $\operatorname{id}_d$.


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