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 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 $X_{/f}$.


1 Answer 1


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.


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