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I'm learning the concept of algebras over $\infty$-operads, following Higher Algebra. The simplest case is when the operad being the trivial operad $\mathrm{Triv}^\otimes$, defined as the 1-full subcategory of the category of finite marked sets $\mathrm{Fin}_*$ containing only inert morphisms. For a $\mathrm{Triv}^\otimes$-monoidal $\infty$-category $\mathcal{C}^\otimes$, the $\infty$-category $\mathrm{Alg}_{\mathrm{Triv}}(\mathcal{C})$ of $\mathrm{Triv}^\otimes$ algebras in $\mathcal{C}^\otimes$ is defined as the full subcategory of $\mathrm{Fun}_{\mathrm{Fin}_*}(\mathrm{Triv}^\otimes,\mathcal{C}^\otimes)$ consisting of functors that preserving inert morphisms. In Higher Algebra 2.1.3.5 it is asserted that evaluation at $\langle 1 \rangle \in \mathrm{Triv}^\otimes$ produces an equivalence $$ \mathrm{Alg}_{\mathrm{Triv}}(\mathcal{C}) \xrightarrow{\sim} \mathcal{C}, $$ which is proven using $p$-Kan extensions, which I'm not familiar with.

I tried to work out this equivalence by myself. By definition $\mathrm{Alg}_{\mathrm{Triv}}(\mathcal{C})$ is the category $\mathrm{Fun}_{/\mathrm{Triv}^\otimes}(\mathrm{Triv}^\otimes,\mathcal{C}^\otimes)^{\mathrm{CCart}}$ of cocartesian sections of $\mathcal{C}^\otimes \to \mathrm{Triv}^\otimes$, which can be further identified with the categorical limit of a diagram $p:\mathrm{Triv}^\otimes \to \mathrm{Cat}_\infty$. By definition we can identify this diagram as $$ p(\langle n \rangle) = \mathcal{C}^{\times n} $$ with the inert morphisms in $\mathrm{Triv}^\otimes$ mapping to projection functors.

However, I have trouble identifying this limit with $\mathcal{C}$, for there are too much interwining arrows. For example, there are 6 arrows in $ \mathrm{Triv}^\otimes$ from $\langle 3 \rangle$ to $\langle 2 \rangle$! My idea is to try transforming this limit diagram in some way, but I did not find a suitable way. Is this approach possible?

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One can avoid relative Kan extensions, but since this calculation hinges on the fact that $\mathcal{C}^\otimes$ is right Kan extended from $\{\langle 1 \rangle\} \to \mathrm{Triv}^\otimes$, I suspect some of the theory of Kan extensions will be necessary to juggle these limits. In particular, this transformation relies not so much on the properties of $\mathrm{Triv}^\otimes$ as the properties of $\mathcal{C}^\otimes$, so I don't expect clever manipulations of $\mathrm{Triv}^\otimes$ to suffice. However, since pointwise Kan extensions amount to requiring various things be limit diagrams, one can at least unfold the argument to see how the limits manipulated more explicitly.

The requirement $\mathcal{C}^\otimes$ is an $\mathrm{Triv}^\otimes$-monoidal amounts to the assumption that $\mathcal{C} : \mathrm{Triv}^\otimes \to \mathrm{Cat}$ is defined as $\langle n \rangle \mapsto \lim_{\langle 1 \rangle \rightarrowtail \langle n \rangle} \mathcal{C}$. This captures not only that the fibers are products and morphisms are projections, but all the compositions and higher coherences come from this. With this observation to hand, the computation becomes possible:

$$ \begin{align*} &\lim\nolimits_{\langle n \rangle : \mathrm{Triv}^\otimes} \lim\nolimits_{f : \langle 1 \rangle \rightarrowtail \langle n \rangle} \mathcal{C} \\ &\cong \lim\nolimits_{(\langle n \rangle, f) : \mathrm{Triv}^\otimes_{\langle 1 \rangle/}} \mathcal{C} && \text{Lim over base/fibers of a left fib = lim over total space} \\ &\cong \lim\nolimits_{\{\langle 1 \rangle\}} \mathcal{C} && \text{Coslice categories have an initial object.} \\ &\cong \mathcal{C} \end{align*} $$

If one couches this in the language of Kan extensions, we note that $\lim_{\mathcal{D}} : \mathrm{Cat}^{\mathcal{D}} \to \mathrm{Cat}$ is (by definition) right Kan extension along $\mathcal{D} \to \mathbf{1}$. The above proof is then a special case of the general fact that Kan extensions compose:

$$ \lim\nolimits_{\mathrm{Triv}^\otimes} \mathcal{C}^\otimes = \mathrm{ran}_{\mathrm{Triv}^\otimes \to \mathbf{1}}\, \mathrm{ran}_{\langle 1 \rangle \to \mathrm{Triv}^\otimes}\, \mathcal{C} = \mathrm{ran}_{\mathbf{1} \to \mathbf{1}}\, \mathcal{C} = \mathcal{C} $$

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