# ($1$-)pullbacks of Kan complexes

Ultimately, I'm trying to figure out whether or not the full subcategory in $$\mathbf{sSet}$$ spanned by Kan complexes is finitely complete (as a $$1$$-category). Since fibrations are stable under pullback in general, I know that Kan complexes are closed under finite products, so the question boils down to whether the pullback in the square $$\require{AMScd}$$ $$\begin{CD} K\times_LK' @>>> K\\ @VVV @VVV\\ K' @>>> L \end{CD}$$ where $$K$$, $$K'$$, and $$L$$ are all Kan complexes must have $$K\times_LK'$$ as a Kan complex also. In my limited experience, I feel like this isn't true since it's not true in a general model category, but I can't construct a counterexample.

I'm honestly pretty bad at creating Kan complexes in general, and my usual go-to's (simplicial groups and nerves of groupoids) are actually preserved under taking pullbacks (the former because $$\mathbf{Grp}$$ is complete and limits of simplicial sets/groups are computed levelwise; the latter because the nerve is fully faithful from $$\mathbf{Cat}$$ to $$\mathbf{sSet}$$ and pullbacks of groupoids are groupoids). Maybe my intuition is wrong?

• It's probably better to try to prove the more general result: Given a commutative cube in $sSet$ where the vertical edges are Kan fibrations except for the 'top left corner' one, and the top and bottom squares are pullbacks, can you prove the last vertical edge is a Kan fibration? That said, I don't know if this is actually true, but it seems that if the special case were true, the more general would be, and likewise if they are false. – David Roberts Jul 9 '20 at 5:06

Take any simplicial set $$X$$ which is not a Kan complex. Let $$K$$ be a Kan replacement of $$X$$, and let $$L$$ be a Kan replacement of the pushout $$K\amalg_X K$$. Then the two maps $$K\to L$$ are levelwise injective, and the pullback $$K\times_L K$$ is precisely $$X$$.