# Lurie's reformulation of symmetric monoidal tensor categories in HA

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 $$\operatorname{id}_d$$.