6
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Is there a particular kind of application you have in mind? $\endgroup$ Mar 21 '10 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 '10 at 13:47
7
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.