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We can define the (first) homology of a surface $S$ by working with graphs embedded in $S$. That is, we take any (oriented) graph which is 2-cell embedded in $S$, and take cycles modulo boundaries in the usual way. Here, I am talking about homology with coefficients in $\mathbb{Z}$. The group that we get is independent of the graph, so is indeed a topological invariant of the surface.

I work with group-labelled graphs, which are oriented graphs with their edges labelled from a finite abelian group $\Gamma$. Proceeding as above, group-labelled graphs allow us to define group-labelled surfaces. That is, let $G$ be a $\Gamma$-labelled graph 2-cell embedded in a surface $S$. If each face of the embedding has group-value zero (the labels of edges on the boundary of the face sum to zero), then this gives a well-defined map on homology. In fact, the embedding of $G$ in S induces a homomorphism from $H_1(S)$ to $\Gamma$. So, we can forget about the $\Gamma$-labelled graph and just study this homomorphism.

My question is: how does this construction relate to taking homology with coefficients from $\Gamma$?

Someone once told me that what I am really doing is working with cohomology with coefficients in $\Gamma$, but I didn't really get this. Can someone please clarify?

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Depending on how you make your labelling precise it sounds like you're either talking about plain-old cellular cohomology with coefficients (see for example Bredon or Hatcher's books), or you're perhaps taking a further step and looking at cellular cohomology with local coefficients. Hatcher's book covers both ideas in detail. –  Ryan Budney Mar 18 '10 at 22:52
    
Sorry if I was imprecise, but the only restriction on the group-labelling is that each face has group-value zero. I guess my question is that there are two ways to 'mod out' by the group $\Gamma$. One is to work with $\Gamma$-labelled graphs satisfying the face condition. This leads to homomorphisms from $H_1(S)$ into $\Gamma$. The second is to work with coefficients in $\Gamma$, either in homology or cohomology. I was wondering how the two notions are related. Thanks for the links everyone. –  Tony Huynh Mar 18 '10 at 23:11
    
Pardon my ignorance, what does "2-cell embedded" mean? –  K.J. Moi Mar 18 '10 at 23:22
    
2-cell embedded means that every face is a disk. –  Tony Huynh Mar 18 '10 at 23:23
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Hi Tony.

This is not really a homology-question, the core of it is the fundamental group. The homomorphism you are using is used in the study of Van Kampen diagrams. Consider a presentation $G=\langle A|R\rangle$. A Van Kampen diagram on $S$ is a labeled graph like you have defined it. The only difference is that in a Van Kampen diagram all labels are generators (or their inverses) $a^{\pm 1}$ and not arbitrary words (although you could define it in this general way without problems because of the Van Kampen lemma).

Then every path in this graph has a group word written on it and "reading the word along a path" is a homomorphism {Paths}$\to G$ with respect to composition of paths. It turns out, that this is compatible with homotopy of paths so this induces a homomorphism $\pi_1(S,x_0)\to G$.

This is the general version of your homomorphism: If your $G$ happens to be abelian, then this homomorphism factorizes through $\pi_1(S,x_0)^{ab}$ which is $H_1(S)$ by the Hurewicz theorem.

This point of view clarifies some connections between the geometry of Van Kampen diagrams and group theoretic questions.

For example the Van Kampen lemma tells you that a group word is trivial if and only if there is a Van Kampen diagram on this disk with this word written on the boundary.

Another fact is this one: If there are no nontrivial "reduced" Van Kampen diagrams on the torus, then every two commuting elements of $G$ generate a cyclic subgroup (i.e. $xyx^{-1}y^{-1}=1$ has only the trivial solutions $x=a^k, y=a^m$ for some $a\in G$.). In a similar spirit one can prove: If there are no nontrivial reduced Van Kampen diagrams on the real projective plane, then there are no involutions in $G$ (i.e. $x^2=1$ has only the trivial solution $x=1$), and if there are no nontrivial reduced Van Kampen diagrams on Klein's bottle, then the only element that is conjugated to its own inverse is the identity (i.e. $yxy^{-1}=x^{-1}$ has only the trivial solution $x=1$).

This connection between geometry and group properties becomes less obscure, if one knows the fundamental groups of the disk (1), the torus ($\langle x,y | xyx^{-1}y^{-1}=1\rangle$), the real projective plane ($\langle x | x^2=1\rangle$) and Klein's bottle ($\langle x,y | yxy^{-1}=x^{-1}\rangle$).

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Hi Johannes. Thanks for the response. I was wondering if this is indeed a question about homotopy rather than homology. For example, if I take $\Gamma=\pi_1 (S, x_0)$, then there should be a $\Gamma$-labelling of the graph $G$ so that each cycle in $G$ containing $x_0$ has group-value precisely its homotopy class. I can see this if I take $\Gamma =H_1(S)$, since I can just label each edge of a spanning tree $T$ of $G$ zero, and each non-tree edge $e$ with the homology class of the unique cycle in $T \cup e$. Since these cycles generate the homology, this works. –  Tony Huynh Mar 19 '10 at 5:23
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Browsing a good introduction to algebraic topology up to the Universal coefficient theorem for cohomology would be a good plan. Hatcher's book, which you can get online, for example.

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May's book is also available online on his website if you're not as interested in the geometric aspects of AT. –  Harry Gindi Mar 18 '10 at 22:49
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The reason that this is cohomology and not homology is that you are looking at functions from the cells to the group.

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Perhaps I should be more explicit about the function that I am describing. It is a function $h$ from the closed walks (cycles) of the group-labelled graph $G$ into the group $\Gamma$. Explicitly, $h$ maps a cycle $C$ to the (oriented) sum of the group-labels of the edges of $C$. With the condition that all faces of $G$ have group-value zero, this map becomes well-defined on homology. Let $C_1$ and $C_2$ be homologous. Then, $C_1 \cup C_2$ is the boundary of an orientable component of $S - C_1 \cup C_2$. Since each face of $G$ has group-value zero, we have $h(C_1)-h(C_2)=0$, as required. –  Tony Huynh Mar 21 '10 at 21:00
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