# On equivalences of cartesian fibrations

Let $$p:X^{\natural} \to S$$, $$q:Y^{\natural} \to S$$ be two cartesian fibrations of simplicial sets (where the marking is given by cartesian edges) and assume that we are given an equivalence of cartesian fibrations $$X^{\natural} \to Y^{\natural}$$.

Given a marking on $$S$$ we define the marked simplicial set $$X^{\dagger}$$ (resp. $$Y^{\dagger}$$) by declaring and edge to be marked if and only if it is marked in $$X^{\natural}$$ and its image under $$p$$ is marked in $$S$$.

My question is the following: Is the induced map $$X^{\dagger} \to Y^{\dagger}$$ a weak equivalence of marked simplicial sets (over the point)?

• What is a marking? – Gerrit Begher Mar 11 '20 at 14:11
• @GerritBegher a choice of 1-simplices that contains all degenerate edges. – F.Abellan Mar 11 '20 at 14:14

Yes. Since $$X^{\natural} \to S$$ and $$Y^{\natural} \to S$$ are both cartesian fibrations they are fibrant and cofibrant objects in the cartesian model structure over $$S$$, which is a simplicial model structure. If two such objects are weakly equivalent then there must exist maps $$f:X^{\natural} \to Y^{\natural}$$ and $$g: Y^{\natural} \to X^{\natural}$$ over $$S$$ with homotopies $$\eta:(\Delta^1)^{\sharp} \times X^{\natural} \to X^{\natural}$$ and $$\tau:(\Delta^1)^{\sharp} \times Y^{\natural} \to Y^{\natural}$$ (again over $$S$$) from $$g \circ f$$ and $$f \circ g$$ to the respective identities. These maps and homotopies imply in particular that $$X^{\natural}$$ and $$Y^{\natural}$$ are equivalent as marked simplicial sets over the point. This remains true if we remove some of the marking, as long as we do it in a way that depends only on some marking of $$S$$. Indeed, as long as $$f,g,\eta$$ and $$\tau$$ remain marking-preserving they will continue to determine an equivalence.