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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)?

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

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