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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:

  1. 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 $MCG(F)$.

  2. how point 1 should give a relation between the $MCG$ and the topological type of $E$

  3. 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!

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maybe – yoyo Jun 30 '11 at 15:53
Unfortunately it does not deal with that stuff. Thanks, anyway. – fatoddsun Jun 30 '11 at 20:24

Just to leave answers as answers:

There is a new 500$+$ page book by Benson Farb and Dan Margalit called A Primer on Mapping Class Groups. I am not an expert in this area, but I'm not the polar opposite of one either, and the book looks to me to be very strong and likely to become a standard reference. So anyone who is at all interested in this subject would do well to download a copy of the book from the second author's webpage while free copies are still available (I have no reason to believe that a copy of this version of the book will go offline once the final version is published, but it doesn't hurt to be safe).

Added: the OP has made clear that his specific questions are not addressed in this book, so this is a very partial answer (more of an answer to the title than the question itself). I hope others will contribute more pertinent information.

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Apparently your answer doesn't answer the question, as the OP left a comment indicating that his questions aren't addressed in that book. – t.b. Jun 30 '11 at 20:47
Indeed, at least as far as I know, there's just a small part regarding surface bundles, but it's not enough to cover the three points that I have indicated. Anyway, it is a very good book..! – fatoddsun Jun 30 '11 at 22:12

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.

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