TITLE: Why does (Ribbon) Graph (co)Homology Compute (co)Homology of MCG?
QUESTION [13 upvotes]: The title says it all.  I am looking for an explanation or reference for why the homology of the ribbon graph complex computes the cohomology of the mapping class groups of surfaces.
I've seen explanations of this using operads, but my understanding is that the operad viewpoint is more recent and not how the above question was originally understood.

REPLY [11 votes]: My favorite way to 'see' the connection between ribbon graphs and mapping class groups is to use the contractibility of the complex of arcs in a surface.  Given a surface $S$, the arc complex $\mathcal{A}(S)$ has vertices given by isotopy classes of arcs with endpoints lying on the boundary of the surface, and simplices given by disjoints collections of such arcs.  Hatcher has a rather attractive way to construct a contraction of this complex down to a single vertex.
By an easy inductive argument one can deduce from the contractibility of the arc complex that the poset $\mathcal{A}_0(S)$ of simplices for which the arcs cut the surface into pieces that are all discs is also contractible.  Such a collection of arcs has a dual ribbon graph.
The mapping class group $MCG(S)$ acts on the poset $\mathcal{A}_0(S)$, and the homotopy quotient is equivalent to the classifying space of the mapping class group since the poset is contractible.  On the other hand, one can build a model for the homotopy quotient by taking the category $MCG(S) \int \mathcal{A}_0(S)$ in which objects are objects of $\mathcal{A}_0(S)$ and a morphism $x \to y$ is a mapping class group element $\alpha$ such that $\alpha\cdot x \subset y$ (it is a theorem of Thomason that this models the homotopy quotient).  By sending a collection of arcs to its dual ribbon graph one sees that this category is equivalent to the category of ribbon graphs of type $S$.