# ($1$-)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. Jul 9, 2020 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 $$X$$.