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 $X_{/f}$.


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.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.