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Let $p:X \to S$ be a right fibration of simplicial sets; one can roughly think of it as some sort of "functor" $S^{op} \to Set_\Delta$ (where $Set_\Delta$ denotes simplicial sets) sending $s \in S$ to the fiber $X_s=p^{-1}(s)$ which is a Kan complex. Functor is in quotes because the problem is this isn't actual functor, i.e. for an edge $s \to s'$, you would need to define a map of simplicial sets $X_{s'} \to X_s$, but it is not clear how...you could also instead of trying to define a simplicial map use the correspondence $p^{-1}(s, s')$ but composition of correspondences gives some trouble. If I understand correctly (and I'm not sure I do, so please correct me), to construct an actual functor $F:\mathfrak{C}[S]^{op} \to Set_\Delta$ such that $F(s)$ is weak homotopy equivalent to $X_s$, Lurie in Higher Topos Theory 2.2.1 defines a straightening functor $St(X):\mathfrak{C}[S]^{op} \to Set_\Delta$ from the simplicial category $\mathfrak{C}[S]^{op}$ built out of $S$ to $Set_\Delta$ as follows: let $$\mathcal{M}=\mathfrak{C}[S] \underset{\mathfrak{C}[X]}{\coprod}\mathfrak{C}[X^\triangleright]$$ then for $s \in \mathfrak{C}[S]$, define $$St(X)(s)=Map_{\mathcal{M}}(s, \infty)$$ where $\infty$ is the cone point of $X^\triangleright$.

My question is why does this definition give what we want? In particular, is $Map_{\mathcal{M}}(s, \infty)$ weak homotopy equivalent to $X_s$? We also need $St X$ to be a simplicial functor and I suppose this comes about because $\mathcal{M}$ is a simplicial category, there is a simplicial map $$Map_{\mathcal{M}}(s, s') \times Map_{\mathcal{M}}(s', \infty) \to Map_{\mathcal{M}}(s, \infty)$$

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    $\begingroup$ Have you taken Lurie's definition apart in the case in which $S$ is a quasi-category and in particular when it is the homotopy coherent nerve of a small simplicial category? I do not, off hand know the answer to your question but that may allow you to interpret things in a more categorical way. $\endgroup$
    – Tim Porter
    Nov 21, 2014 at 14:56
  • $\begingroup$ In section 2.2.2 (which I hadn't read when I asked this) Lurie works out the case when S is a point, and then Prop. 2.2.2.7 seems to answer my question. $\endgroup$
    – ykm
    Nov 22, 2014 at 8:39

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