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?

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    $\begingroup$ 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. $\endgroup$ Jul 9, 2020 at 5:06

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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 $X$.


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