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.