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](http://www.math.washington.edu/~warner/TTHT_Warner.pdf) 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?



Homotopy pullbacks of simplicial spaces, and Bousfield-Friedlander
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.


