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I have read many proofs of this theorem (topological) but all of them use the idea of graphs or euler characteristic. I was wondering if there is a topological proof using covering spaces but without these 2 ideas.

More precisely, I know that every subgroup H of index k of a free group G corresponds to a k-sheeted covering space (E,e) such that H=$p_{*}(\pi_1(E,e))$. It is possible to be able to draw $\pi_1(E,e)$ in some cases and hence deduce that $\pi_1(E,e)$ is free with k(n-1)+1 generators (example k=2 and n=2). But my question is if I know that $\pi_1(E,e)$ is free with 3 generators for example, how can I deduce that $H$ has 3 generators? All we have is that $H=p_{*}(\pi_1(E,e))$.

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  • $\begingroup$ It's not really clear to me what you want, because you have not specified any specific topological spaces, and you have in fact ruled out graphs in your proof, and yet you say you want a topological proof. $\endgroup$
    – Lee Mosher
    Commented Apr 27, 2020 at 20:53
  • $\begingroup$ For example in the case of k=2 and n=2, I just draw the topological space without mentioning that is a graph (I use the topological space for a circle for instance) I get exactly what you call the graph (I verify it in hatcher's book in his illustrations). So what i mean is: noticing that the fundamental group of E has 3 generators, is there anyway I can deduce that its image by p_{*} has also 3 generators? $\endgroup$
    – user752801
    Commented Apr 27, 2020 at 20:55
  • $\begingroup$ Maybe using the injectivity of p_{*}? $\endgroup$
    – user752801
    Commented Apr 27, 2020 at 20:56
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    $\begingroup$ Yes there is: the homomrophism $p_*$ is injective (because $p$ is a covering map), and therefore $\pi_1(E,e)$ is isomorphic to $p_*(\pi_1(E,e)$; and to be free with 3 generators is a group property that is invariant under isomorphism. $\endgroup$
    – Lee Mosher
    Commented Apr 27, 2020 at 20:56
  • $\begingroup$ Oh perfect. That is exactly what i need! $\endgroup$
    – user752801
    Commented Apr 27, 2020 at 20:57

1 Answer 1

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The homomorphism $p_*$ is injective (because $p$ is a covering map), and therefore $\pi_1(E,e)$ is isomorphic to $p_*(\pi_1(E,e))$. And, to be free with some particular number $n$ of generators is a property of groups that is invariant under isomorphism, so assuming this property holds for $\pi_1(E,e)$, it also holds for $H = p_*(\pi_1(E,e))$.

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