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I am using a definition of the first homology group by Miranda (Algebraic Curves and Riemann surfaces), which is as follows,

The fundamental group $\pi_1(X,\alpha)$ is the group that consists of homotopy classes of closed paths starting and ending at $\alpha \in X$. Let $[\pi_1,\pi_1]$ be its commutator subgroup. We define the first homology group of $X$ to be $H_1(X)=\pi_1(X,\alpha)/[\pi_1,\pi_1]$.

I am wondering now, is in this definition $H_1(X)$ the same as the integral homology $H_1(X,\mathbb{Z})$?

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    $\begingroup$ Yes, it is..... $\endgroup$ – Randall May 17 '18 at 11:42
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This is not the same definition, of course, but the result is isomorphic to the usual definition if $X$ is path-connected. This is called the Hurewicz theorem.

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  • $\begingroup$ For a 'reference' : en.wikipedia.org/wiki/Hurewicz_theorem $\endgroup$ – peter a g May 17 '18 at 12:22
  • $\begingroup$ @TheBeiram : here is an actual (presumably correct!) reference: page 166 of math.cornell.edu/~hatcher/AT/AT.pdf $\endgroup$ – peter a g May 17 '18 at 12:31
  • $\begingroup$ Thank you. I can't really see the intuitive difference in the definitions of $H_1(X)$ and $H_1(X,\mathbb{Z})$, can you help me with that? $\endgroup$ – TheBeiram May 17 '18 at 12:45
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    $\begingroup$ @TheBeiram What do you mean you cannot see the difference? In one case you look at the group of homotopy classes of loops and you abelianize it, in the other case you look at formal sums of paths, you consider the cycles, and you mod out by the boundaries. The definitions are obviously different, that's why it's a theorem that they give the same result. $\endgroup$ – Najib Idrissi May 17 '18 at 12:49

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