Let $X_\bullet \longrightarrow Y_\bullet \longleftarrow Z_\bullet$ be a diagram of simplicial spaces (=bisimplicial sets, if you like).

On p. 14-9 of these notes there is an example which shows that if $Y_\bullet$ and $Z_\bullet$ are levelwise connected then the homotopy pullback of the geometric realisation of the diagram is the geometric realisation of the levelwise homotopy pullback.

The theorem is proved using the Bousfield-Friedlander theorem, which only requires that $\pi_0Z_\bullet \to \pi_0Y_\bullet$ is a Kan fibration and that $Z_\bullet$ and $Y_\bullet$ are $\pi_*$-Kan. Being levelwise connected implies both of these conditions, but is not necessary.

Can the conditions of the Bousfield-Friedlander theorem be relaxed? How about if $Z_\bullet \to Y_\bullet$ is something like a $\pi_*$-Kan fibration", though I'm not sure of the precise definition this should have?

  • $\begingroup$ Is there a particular kind of application you have in mind? $\endgroup$ Mar 21, 2010 at 13:42
  • $\begingroup$ Well, for my purposes the simplicial spaces would be nerves of topological monoids (not group-like!), which in turn arise as the nerves of certain monoidal groupoids. $\endgroup$ Mar 21, 2010 at 13:47

1 Answer 1


The following represents what I know about this; I don't know of a published reference.

Given a map $f:Z_\bullet\to Y_\bullet$ of simplicial "spaces" (to make things easy, assume spaces are simplicial sets), let's call it a realization quasi-fibration (RQF) if for every $U_\bullet\to Y_\bullet$, the homotopy pullback of the geometric realizations is weakly equivalent to the realizations of the levelwise homotopy pullbacks. The Bousfield-Friendlander theorem gives a sufficient condition for $f$ to be a RQF, in terms of the dreaded $\pi_*$-Kan condition.

Some facts:

  • The pullback of an RQF $f$ along any $U_\bullet\to Y_\bullet$ is itself an RQF.

  • Let $F[n]_{\bullet}$ be the simplicial space which is free on a point in degree $n$. Then $f$ is an RQF if and only if its pullback along all $g: F[n]_\bullet \to Y_\bullet$, for all $n$, is an RQF.

These two facts are consequences of something that is sometimes called "descent"; basically, the facts that homotopy colimits distribute over homotopy pullbacks, and compatible homotopy pullbacks assembled by a homotopy colimit result in a homotopy pullback.

So the above gives exact criteria for $f$ to be an RQF. Whether the pullback of an RQF $f$ along a map $g$ is again an RQF only depends on the homotopy class of $g$. So if $f:Z_\bullet\to Y_\bullet$ is any map, let $\pi_0Y$ be the simplicial set whose $k$-simplices are $\pi_0(Y_k)$, which is to say all homotopy classes of maps $F[k]_\bullet\to Y_\bullet$. Let $RQF(f)\subseteq \pi_0Y$ be the sub-simplicial set whose $k$-simplices correspond to $g:F[k]_\bullet\to Y_\bullet$ such that the pullback of $f$ along $g$ is an RQF.

So the criterion is: $f$ is an RQF iff $RQF(f)=\pi_0Y$.

It turns out that since geometric realization always preserves products, any map $Z_\bullet \to point_\bullet$ is an RQF. Thus $RQF(f)$ contains all $0$-simplices of $\pi_0Y$. Thus, if all $Y_k$ are connected, $f$ is an RQF, which implies the result you describe.


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