TITLE: Article or book explaining rigorously facts about the mapping class group
QUESTION [5 upvotes]: I would like to know more about relationships between the mapping class group of an orientable surface with negative Euler's characteristic and moduli spaces.
In particular, I would like to have a rigorous explanation of the following three facts:

having a family of surfaces given by a fibration $F\to E\to B$ (e.g. take B as $S^1$ and F as a compact surface of genus $g>1$), how to construct a map from $\pi_1(B)$ to $MCG(F)$.
how point 1 should give a relation between the $MCG$ and the topological type of $E$
how the cohomology of the $MCG$ is related with the cohomology of the moduli space of curves

In other words, I know just some chatting about that, but I would like to learn more deeply how to formulate and prove those facts.
Thanks a lot in advance,
bye!

REPLY [2 votes]: This is an old question, but it deserves a reasonable answer. 
For Question 1:
Assuming that $F$ has genus $\ge 2$, isomorphism classes of $F$-bundles over any (smooth) manifold $M$ (e.g. the circle) are in a natural bijective correspondence with conjugacy classed of homomorphisms $\pi_1(M)\to MCG(F)$. 
See Proposition 4.6 in S.Morita, "Geometry of Characteristic Classes". When $F$ has genus $\le 1$ the situation is a bit different, see the discussion in Morita's book preceding Proposition 4.6. 
This answers your question 1 in the case of a (connected) manifold base. When the base is not a manifold, but is, say, a connected finite-dimensional CW complex $B$ one can prove the same thing by replacing  $B$ with a (noncompact) smooth manifold homotopy equivalent to $B$. (I am not sure if you really want to see a proof since the question is quite old.) 
As for Question 2, I am not sure how much do you want to know here. If total spaces of two fibrations are homeomorphic, the homeomorphism need not preserve the fibrations and things become quite complicated very quickly.
For Question 3: It depends on what do you mean by "cohomology". If you work with, say, rational, coefficients (or ${\mathbb Z}/p$ coefficients, $p$ is sufficiently large), then indeed 
$$
H^*(MCG(F), {\mathbb Q})\cong H^*({\mathcal M}_g, {\mathbb Q})
$$ 
where $g$ is the genus of $F$ and ${\mathcal M}_g$ is the moduli space of genus $g$ surfaces. This is because ${\mathcal M}_g$, treated as a DM stack (aka an orbifold), has contractible universal cover and the fundamental group isomorphic to $MCG(F)$. See page 146 of Morita's book. 
This is a special case of a more general theorem about non-free properly discontinuous group actions on CW complexes, you should be able to find this in Brown's book "Cohomology of groups". But if you work, say, with integer coefficients, then things are more complicated.