Associated fibration to a representable infinity functor

Question

Let $p:X\rightarrow S$ be a right fibration of simplicial sets. Lurie defines in Higher Topos Theory Definition 3.3.2.2. that a funtor $f:S \rightarrow \mathcal{S}$ with $\mathcal{S}$ the infinity category of spaces classifies p if there exisists an equivalence of right fibrations $X\rightarrow Un_{S}(f)$.
I would like to show that for the case f = j(s), where j denotes the infinity categorical Yoneda embedding and s a vertex of S, f represents the right fibration $S_{s/} \rightarrow S$.
Could someone show me please how the calculation goes?

Answer 1

I'll write $\mathcal C$ instead of $S$ to avoid confusion with the $\infty$-category $\mathcal S$ of spaces. Also $\mathcal C_{s/}\rightarrow \mathcal C$ is a left fibration but no right fibration in general (I assume that's just a typo), so I'll work with $\mathcal C_{/s}\rightarrow \mathcal C$ instead, as probably intended.

The proof you asked for follows from Higher Algebra, Section 5.2.1, although the argument is rather indirect. Lurie first defines a right fibration $\operatorname{TwArr}(\mathcal C)\rightarrow \mathcal C\times \mathcal C^{\mathrm{op}}$ called the "twisted arrow category" of $\mathcal C$. Via straightening, this defines a functor $\chi\colon \mathcal C\times \mathcal C^{\mathrm{op}}\rightarrow \mathcal S$ into the $\infty$-category $\mathcal S$ of spaces. In the proof of Proposition 5.2.1.10 it is checked that the pullback of $\operatorname{TwArr}(\mathcal C)\rightarrow \mathcal C\times \mathcal C^{\mathrm{op}}$ to $\mathcal C\times\{s\}$ is equivalent to the right fibration $\mathcal C_{/s}\rightarrow \mathcal C$ for every 0-simplex $s\in \mathcal C^{\mathrm{op}}$, so the restriction of $\chi$ to $\mathcal C\times\{s\}$ is equivalent to the functor $\mathcal C^{\mathrm{op}}\rightarrow \mathcal S$ obtained as the straightening of $\mathcal C_{/s}\rightarrow \mathcal C$. In Proposition 5.2.1.11 Lurie proves that the functor $\mathcal C\rightarrow \operatorname{Fun}(\mathcal C^{\mathrm{op}},\mathcal S)$ induced by $\chi$ is equivalent to the Yoneda embedding $j\colon \mathcal C\rightarrow \operatorname{Fun}(\mathcal C^{\mathrm{op}},\mathcal S)$ from Higher Topos Theory, Section 5.1.3. Combining these two facts shows that $j(s)$ is indeed given by the straightening of $\mathcal C_{/s}\rightarrow \mathcal C$.