In HTT, given a inner fibration $p : X \rightarrow S$ of simplicial, an edge $f : x \rightarrow y$ of the simplicial set $X$ is said to be a $p$-Cartesian if the induced map 
$$ X\_{/f} \rightarrow X\_{/y} \times\_{S\_{/p(y)}} S\_{/p(f)}$$ is a trivial Kan fibration.


In Remark 2.4.1.4 Lurie says that this definition is equivalent to the following one : in the same setting, $f$ is $p$-Cartesian if and only if for every $n \geq 2$ and every commutative diagram


\begin{matrix} \Delta^{\{n-1,n\}} \\
\downarrow \scriptstyle&&\\
\Lambda^n\_n&{\to}&X\\
\downarrow &&\downarrow \scriptstyle{p}\\
 \Delta^n&{\to}&S\end{matrix}
where the composition $\Delta^{\{n-1,n\}} \rightarrow \Lambda^n\_n \rightarrow X$ is the edge $f$ ( I don't know how to make a "bottom right" arrow) there is a map $h : \Delta^n \rightarrow X$ rendering the diagram commutative.


I don't see why those definitions are equivalent. For example if we have the first one, given a diagram
\begin{matrix} \Delta^{\{n-1,n\}} \\
\downarrow\scriptstyle&&\\
\Lambda^n\_n &\stackrel{\alpha}{\to}&X\\
\downarrow & &\downarrow \scriptstyle{p}\\
 \Delta^n&\stackrel{\beta}{\to}&S\end{matrix}


we have a map $ \Delta^{n-2} \rightarrow S\_{/p(f)}$ using $\beta$ and the universal property of the slice but I don't see how to get a map from $\Delta^{n-2}$ into $X\_{/y}$ nor from a simplicial subset of $\Delta^{n-2}$ into $X\_{/f}$.



Remark 2.4.1.4 Higher Topos Theory
Let's see what the data of a diagram


\begin{matrix}
\partial\Delta^{n-2}&{\to}&X\_{/f}\\
\downarrow &&\downarrow\\
 \Delta^{n-2}&{\to}&X\_{/y}\times\_{S\_{/p(y)}}S\_{p(f)}\end{matrix}


translates to. I claim this is formally the same as a diagram


\begin{matrix}
\partial\Delta^{n-2}\star\Delta^{\{n-1,n\}}\sqcup\_{\partial\Delta^{n-2}\star\Delta^{\{n\}}}\Delta^{n-2}\star\Delta^{\{n\}}&{\to}&X\\
\downarrow &&\downarrow\\
 \Delta^{n-2}\star\Delta^{\{n-1,n\}}&{\to}&S\end{matrix}


and that the data of a lift is the same in both diagrams. This currying-like pattern appears over and over again in HTT - maybe the place where it is spelled out most explicitly is in Remark A.3.1.6. (which is for mapping spaces, but essentially the same thing applies for under- or over-categories.)


Anyways, the left hand arrow coincides with $\Lambda^n\_n\rightarrow\Delta^n$, proving the claim.


