Given a sequence of simplicial spaces (actually bisimplicial sets)

$$F\to E\to B$$

that is level-wise a fibration, then the geometric realisation does not necessarily have to be a fibration.

If I understand it correctly then a sufficient criterion is that all spaces are group-complete H-spaces and that the map of simplicial sets $\pi_0(E)\to \pi_0(B)$ is surjective.

Now consider a map of simplicial group complete H-spaces $f:X\to Y$, the canonical maps $p_X:X\to \pi_0(X)$, $p_Y:Y\to \pi_0(Y)$ and the induced map $g:\pi_0X\to \pi_0 Y$.

Let me denote by $\operatorname{hofib} |f|$ the homotopy fibre of the map $f$ and by $|\operatorname{hofib} (f)|$ the geometric realisation of the level-wise homotopy fibres, similarly for $g$.

Then I have induced maps $F: |\operatorname{hofib} (f)|\to \operatorname{hofib} |f|$ and $G:|\operatorname{hofib} (g)|\to \operatorname{hofib} |g|$.

Now $\operatorname{hofib}(F)$ and $\operatorname{hofib}(G)$ measure the defect how far the geometric realisation of the level-wise fibres is away from the actual fibre. By the sufficient condition in the beginning this obstruction should be completely captured by the behaviour on the $\pi_0$. But by construction this is the same for $f$ and $g$. Can I somehow conclude that

$$\operatorname{hofib} (F)\simeq \operatorname{hofib} (G)?$$

Edit: The space $|\operatorname{hofib}(g)|$ as written above doesn't really make sense. In the end I am talking about maps of sets here and over every connected components (i.e. every point) the fibre will be just the preimage of this point. What really should be there is $|\pi_0 \operatorname{hofib}(f)|$. I am not sure if this makes the question easier or harder.

  • 1
    $\begingroup$ It looks to me like "hofib" should be an \operatorname, so I submitted an edit with changed formatting. Hopefully that's what you wanted! $\endgroup$
    – Vectornaut
    Nov 13 '13 at 18:53
  • $\begingroup$ I do not quite understand what exactly is being asked. Nevertheless, this mathoverflow question should be relevant. $\endgroup$ Nov 16 '13 at 19:32

Working with spaces meaning spaces, I gave a reasonably strong result along these lines in Theorem 12.7 of The Geometry of Iterated Loop Spaces http://www.math.uchicago.edu/~may/BOOKS/geom_iter.pdf.

If $p\colon E\to B$ is a simplicial Hurewicz fibration such that $B$ is proper (Reedy cofibrant in more modern terminology) and each $B_q$ is connected, then $|p|\colon |E|\to |B|$ is a quasifibration with fiber $|F|$.

  • $\begingroup$ Yes, I think I have seen that. The problem in my situation is that the $B_q$ are not connected. However I am trying to compare two sequences which have the same $\pi_0E_q$ and $\pi_0B_q$ and was hoping to compare them. Precisely because your result (and others) suggest that the obstruction is captured in $\pi_0B$. $\endgroup$ Nov 14 '13 at 8:49
  • 1
    $\begingroup$ The 1978 survey article ``Fibrations and geometric realizations'' by D.W. Anderson on results of this type might help. $\endgroup$
    – Peter May
    Nov 14 '13 at 13:30

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