I couldn't think of a title for this, but here we go:

Fix $p:S\rightarrow T$, a left fibration of simplicial sets, and an edge $f:\Delta^1 \rightarrow T$. Let $t$ be the first vertex of $f$, and $t'$ be the second vertex.

We name the induced map, $q: S^{\Delta^1}\rightarrow S^{\{1 \}}\times_{T^{\{1 \}}} T^{\Delta^1}$.

Now let $X$ be the simplicial set of ** sections of the projection** $S\times_T \Delta^1\rightarrow \Delta^1$, where the pullback is taken with respect to the map $p:S\rightarrow T$ and the fixed edge $f:\Delta^1 \rightarrow T$.

More notation: We'll denote by $S_{t'}$ the fiber $S \times_T \Delta^0$ where $\Delta^0 \rightarrow T$ is given by the inclusion of the vertex $t'$ and $S\rightarrow T$ is given again by p. We give a fiber $q':X \rightarrow S_{t'}$ of $q$ over the edge $f$ (This is about where I stop understanding what's going on).

What we'd like to show is: $q$ and $q'$ have the same fibers over points of $S^{\{1 \}}\times_{T^{\{1 \}}} T^{\Delta^1}$ where the second projection is the edge f. Remember that exponentiation denotes the internal Hom.

The problem is this simplicial set of sections. What are its maps out, and why do they naturally go to $S_{t'}$ and agree with q? I feel like the key to this is understanding how the exponential is mapping into the pullback, but it's not really clear to me how that should work.

This fact is stated in HTT by Lurie in the proof of proposition 2.1.3.1, but I don't really see how it's obvious.

A link to the relevant proof/page: http://books.google.com/books?id=CTe68E8wK4QC&lpg=PP1&ots=o8qYDiX4mt&dq=lurie%20higher%20topos%20theory&pg=PA67#v=onepage&q=&f=false

Update: "Work" I've done thusfar: $$\ \ \matrix{&S^{\Delta^1}_f &\to & S^{\Delta^1}& \cr &\downarrow &Pb &\downarrow \cr S_{t'}\cong &L_f & \to & L & \to & S^{\{1\}} & \cr &\downarrow &Pb&\downarrow&Pb&\downarrow p \cr &\Delta^{0} & \to & T^{\Delta^1} & \to & T^{\{1\}} \cr &&f&&d_1}\ \ $$

Note that $d_1$ denotes the face map at the vertex 1. Also, $L:= S^{\{1 \}}\times_{T^{\{1 \}}} T^{\Delta^1}$, and The point here is that it should be "morally" the same to give a pullback of $\Delta^1\to T \leftarrow S$ as giving a pullback $\Delta^0\to T^{\Delta^1} \leftarrow S^{\Delta^1}$ with respect to the edge f. So we'd like to show that $X$, the simplicial set of sections of the projection noted before is somehow isomorphic to $S^{\Delta^1}_f$, since this shows that $q$ and $q'$ agree where they're needed to. So what I'm struggling with at this point is showing this last idea.