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The title says it all. I'm still a beginner in homology, learning out of Hatcher, but I've run into mentions of $H_1$ with rational or real coefficients in some papers, and I'm struggling to see how that makes sense in the context of "cycles" in the graph-theoretic sense.

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    I think it already makes no sense to hold on to a picture when the coefficients are in $\mathbb{Z}$. As for "cycles", the terminology merely coincides here. (It is true that a cycle in a graph is also a cycle in the sense of homology. But what does a 2-cycle correspond to? It's better to think of them in terms of simplicial or cellular complexes.)2012-02-29

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The long answer is that for simply connected CW complexes $X$, one can construct another space, $X_{\mathbb Q}$ such that $\widetilde{H}_*(X;\mathbb Q)\cong \widetilde{H}_*(X_{\mathbb Q};\mathbb Z)$ and $\pi_*(X)\otimes\mathbb Q\cong \pi_*(X_{\mathbb Q})$. This construction is called the rationalization of $X$, and is a central construction in rational homotopy theory. (See p.108 of the book Rational Homotopy Theory by Felix, Halperin and Thomas.)

The most basic example is not simply connected but works anyway. It is formed by an infinite telescope, where you take infinitely many copies of $S^1\times I$ lined up in a row, where the left-hand circle from the $n$th copy is attached by wrapping it $n$ times around the right-hand circle of the $n-1$st copy. In other words, let the rational $1$-sphere $S^1_{\mathbb Q}= (S^1\times I)\times \mathbb N/\sim$ where $((\theta,0),n)\sim((n\theta,1),n-1)$. It is an interesting exercise to check that $\pi_1(S^1_{\mathbb Q})\cong \mathbb Q= \pi_1(S^1)\otimes\mathbb Q$.

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    Any topological space with $\pi_1(X)\cong\mathbb R$ would have to be pretty wild, since $\mathbb R$ is uncountable. Such spaces do exist, but I don't know that anybody has studied a functor along the lines of rationalization for $\mathbb R$.2012-03-01