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I am in the context of simplicial sets. I have a square diagram that I want to show to be homotopy pullback. What I can grasp is that it is a pullback up to some equivalences in the middle of my reasoning, and I would like to promote this to a real statement.

So let $p:X \to Z, q:Y \to Z$ map of ssets, and $f:P \to X, g:P\to Y$ other two maps from the supposed-to-be homotopy pullback. Suppose that for every $x \in X$, the induced map on homotopy fibers $P \to Y$ is an equivalence. Furthermore, you can additionally suppose that $p,q$ are Kan fibrations. Is it true that $P$ is the homotopy pullback of the diagram?

Th other point is the following: when the homotopy divers are equivalent to the fibers? Maybe when maximal connected groupoids at the base are contractible?

Related question: Can homotopy pullbacks of spaces be checked on fibers?

The difference is that we are not dealing with spaces (=Kan complexes) but with general simplicial sets. Moreover, I'd like to use actual and not homotopical fibers.

Reference: http://palmer.wellesley.edu/~ivolic/pdf/Papers/CubicalHomotopyTheory.pdf

Proposition 3.3.18. Thanks!

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  • $\begingroup$ Let $X$ be any non-empty fibrant simplicial set with a single vertex and let $P = Y = Z = \Delta^0$. Then fibers of $P$ and $Y$ are always equal to $\Delta^0$, but the square is a homotopy pullback only if $X$ is contractible. $\endgroup$
    – Valery Isaev
    Commented Jun 14, 2019 at 1:43
  • $\begingroup$ Ok, thanks. I will edit the question so that it includes homotopy fibers (seems to solve jour counterexample). $\endgroup$
    – Andrea Marino
    Commented Jun 14, 2019 at 9:40

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