The original question I had was:

If I have a sequence of simplicial spaces

$$A\to B\to C$$

which is degree-wise a homotopy fibration, under which conditions is the geometric realization also a homotopy fibration?

I bet there are tons of results on this. I have found the following theorem published by Anderson:

Theorem

If $X\to Y$ is a map of simplicial spaces such that $\pi_0(f)$ is a Kan-fibration, and if the higher groupoids $\Pi_\infty(X)$ and $\Pi_\infty(Y)$ are fully fibrant, then for any map $g:Y'\to Y$ of simplicial spaces, if $X'$ is the homotopy theoretic fiber product of $Y'$ with $X$ over $Y$, $|X'|$ is the homotopy theoretic fiber product of $|Y'|$ with $|X|$ over $|Y|$.

Now the theorem answers the question, by letting $X=\ast$ and $Y=C$. The condition on $\pi_0(f)$ becomes then something easy, but I am having trouble understanding the motivation behind the $\Pi_\infty(Y)$ condition. In fact I have quite a lot of structure on the simplicial spaces in question and I doubt that it is even prudent to work with the definition itself. Can anybody enlighten me?

For me $C=Y$ is itself in every degree the classifying space of a category and moreover a group-like H-space.

Edit: I just remembered a different result by Waldhausen (Algebraic K-theory of generalized free products, Lemma 5.2)

Let $$A\to B\to C$$ be a sequence of bisimplicial sets such that that the composition is constant. >Suppose geometric realization in one direction gives homotopy fibrations with connected base. Then >the realization in both directions is a homotopy fibration.

So I guess the questions are:

1. Does $C$ being the realization of a connected simplcial set imply that $\Pi_\infty(C)$ is fully fibrant?
2. Can I say something if the base is not connected, but a group-like H-space, i.e. connected components are homotopy equivalent?