If $Y$ is a fibrant simplicial set and $\Delta^{\bullet}$ is the cosimplicial simplicial set, is $Y(\Delta^{\bullet})$ (i.e. $n^{th}$ simplicial set is $n \mapsto Y(\Delta^{n})=Hom(\Delta^{n},Y)$) a (Reedy) fibrant bisimplicial set?
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$\begingroup$ Do you really mean $n\mapsto Hom(Y,\Delta^n)$ or $n\mapsto Hom(\Delta^n,Y)$? Because only the latter is a bisimplicial set. $\endgroup$ – Denis Nardin Apr 19 '18 at 8:15