Let $Y$ be a marked simplicial set, whose underlying simplicial set is also denoted by $Y$. Let $X$ be a scaled simplicial set such that the [decalage](http://ncatlab.org/nlab/show/decalage) of its underlying simplicial set is $Y$. $X$ is then a scaled simplicial set, whose thin $2$-simplices are precisely the marked $1$-simplices of $Y$. 



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> **Question:** If $Y$ is an $(\infty,1)$-category (via the Cartesian model structure on $\mathrm{Set}^+\_\Delta$, see $\S 3.1.3$ of HTT), then is $X$ an $(\infty,2)$-category (via the model structure on $\mathrm{Set}^\mathrm{sc}\_\Delta$, see $\S 4.2$ [here](http://www.math.harvard.edu/~lurie/papers/GoodwillieI.pdf))?
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Building $(\infty,2)$-categories from $\infty$-categories
The following is not an answer, just an observation suggesting that maybe the question should be rephrased. I claim that if $X$ is a fibrant scaled simplicial set whose decalage is a fibrant marked simplcial set then both $X$ and $Y$ must in fact be $\infty$-groupoid (i.e., all the triangles in $X$ are thin, all the edges in $Y$ are marked, and the underlying simplicial sets of both $X$ and $Y$ are Kan complexes). Let me give a proof for the case where $X$ has exactly one vertex $x\_0$ (the proof in the general case is similar). Let $y\_0 \in Y\_0$ be the object corresponding to the degenerate edge $s(x\_0)$. Let $e$ be an edge in $X$ and let $y\_e \in Y\_0$ be the corresponding vertex. Then there is a degenerate triangle in $X$ with edges $s(x\_0),e,e$ whose corresponding edge in $Y$ is not degenerate (recall that in the decalage process the last degeneracy in each dimension disappears). This gives an edge in $Y$ from $y\_0$ to $y\_e$ which must be marked because it corresponds to a degenerate (and hence thin) triangle of $X$. This shows in particular that all the objects in $Y$ are equivalent to each other. Since $Y$ is a fibrant marked simplicial set this means that there must be a marked edge between every two objects in $Y$. This now implies that every map $\Lambda^2\_0 \longrightarrow X$ can be extended to a thin triangle $\Delta^2 \longrightarrow X$, which, in turn, implies that the maximal sub-$(\infty,1)$-category of $X$ is an $\infty$-groupoid (i.e., the maximal sub-simplicial set of $X$ in which all the triangles are thin is a Kan complex). This means that the automorphism category of $x\_0$ is a monoidal $(\infty,1)$-category in which every object has an "inverse object". But it is known that such a monoidal $(\infty,1)$-category must be an $\infty$-groupoid. This implies that $X$ is actually an $(\infty,1)$-category, and hence an $\infty$-groupoid by the above. As a result $Y$ must be an $\infty$-groupoid as well.


**Edit**: I realized the argument above has a problem. It is not true in general that a monoidal $(\infty,1)$-category in which every object has an inverse is an $\infty$-groupoid. An alternative way to finish the argument is as follows. Let $e\_1,e\_2$ be two edges of $X$. Then every three edges in $Y$ of the form $f: y\_0 \to y\_{e\_1}$, $g: y\_{e\_1} \to y\_{e\_2}$ and $h: y\_0 \to y\_{e\_2}$, with $f$ and $h$ marked, can be extended to a triangle $\Delta^2 \to Y$. This is because the corresponding extension problem in $X$ is scaled anodyne (note that the edge of $X$ corresponding to $y\_0$ is degenerate). We may hence conclude that $g$ is marked. But $g$ is completely arbitrary (since marked $f$ and $h$ always exist by the above arguments), and hence all the edges of $Y$ are marked. It follows that all the triangles of $X$ are thin, and hence we may conclude as above that $X$ and $Y$ are both $\infty$-groupoids.


