The effect of straightening on morphisms This is similar to another question on MO, but is different.
Let $p:\mathcal{E}\to\mathcal{C}$ be a right fibration between $\infty$-categories. As explained in Lurie's HTT, $\S$2.1, we can consider $p$ as a functor from $\operatorname{ho}\mathcal{C}^\mathrm{op}$ to the homotopy category of $\mathsf{sSet}_{\mathrm{Quillen}}$: Given a morphism $f:x\to y$ in $\mathcal{C}$, the map $p^{-1}(y)=\mathcal{E}_y\to\mathcal{E}_x$ is defined by restricting a filler in the commutative square
$$
\require{AMScd}
\begin{CD}
\mathcal{E}_y\times \{1\} @>>>\mathcal{E}\\
@VVV @VV{p}V\\
\mathcal{E_y}\times \Delta^1 @>>>\mathcal{C}.
\end{CD}
$$
Equivalently, the map $\mathcal{E}_y\to\mathcal{E}_x$ is obtained by choosing a section of the trivial fibration $\mathcal{E}_y\leftarrow\operatorname{Fun}(\Delta^1,\mathcal{E})\times _{\operatorname{Fun}(\Delta^1,\mathcal{C})}\{f\}$ and then composing the $\operatorname{Fun}(\Delta^1,\mathcal{E})\times _{\operatorname{Fun}(\Delta^1,\mathcal{C})}\{f\}\to\mathcal{E}_x$.
On the other hand, we can apply the straightening functor (HTT, $\S$2.2.1) to obtain a simplicial functor $\operatorname{St}(p):\mathfrak{C}[\mathcal{C}]^\mathrm{op}\to\mathsf{sSet}$. This simplicial functor has the property that for each object $x$ of $\mathcal{C}$, there is a natural zig-zag of weak equivalences between $\mathcal{E}_x$ and $\operatorname{St}(p)(x)$ (HTT, Proposition 2.2.3.15). Just before HTT Proposition 2.1.1.5, there is a comment which connotes that the straightening construction is a refined version of the construction in the previous paragraph. Given that, I expect that, given a morphism $f:x\to y$ in $\mathcal{C}$, the maps $\mathcal{E}_y\to\mathcal{E}_x$ and $\operatorname{St}(p)(y)\to \operatorname{St}(p)(x)$ induced by $f$ are isomorphic in the arrow category of $\operatorname{ho}(\mathsf{sSet}_{\mathrm{Quillen}})$. But I've been unsuccessful at proving this. Is it true, and if so, why?
 A: The answer is yes. The key is the relative nerve functor, discussed in HTT, $\S$ 3.2.5.
We wish to show that if $p:\mathcal{E}\to\mathcal{C}$ is a right
fibration classified by a functor $f:\mathcal{C}^{\mathrm{op}}\to\cal S$,
then given a morphism $\alpha:x\to y$ in $\mathcal{C}$, we can identify
$f(\alpha)$ with the map $\mathcal{E}_{y}\to\mathcal{E}_{x}$
in $\operatorname{ho}(\mathcal{S})$. This is true. Indeed, since
classifying maps are compatible with pullbacks, we may assume that
$\mathcal{C}$ is the nerve of an ordinary category $\mathcal{A}$.
The unstraightening functor $\operatorname{Un}_{\mathcal{C}}:\operatorname{Fun}(\mathcal{C},\mathsf{sSet})_{\mathrm{proj}}\to\mathsf{sSet}/\mathcal{C}_{\mathrm{contra}}$
is naturally weakly equivalent to the (contravariant) relative nerve
functor $r^{*}:\operatorname{Fun}(\mathcal{C},\mathsf{sSet})_{\mathrm{proj}}\to\mathsf{sSet}/\mathcal{C}_{\mathrm{contra}}$
on fibrant objects, so we may assume that $p=r^{*}F$
for some projectively fibrant functor $F:\mathcal{C}\to\mathsf{sSet}$.
By inspection, the map $(r^{*}F)_{y}\to(r^{*}F)_{x}$ associated with
$\alpha$ is $F\alpha$, and we are done.
