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Let $G$, $H$ be free pro-$p$-groups, where $p$ denotes a rational prime.

I want to show that a canonical homomorphism

$$G \to H$$ is in fact an isomorphism. The next step is a reduction step to the abelianizations of $G$, resp. $H$ that I'd like to prove. The claim therefore is

If $G^{ab} \to H^{ab}$ is an isomorphism, we get an isomorphism on the corresponding free pro-$p$-groups $G \to H$.

How does that work?

Thank you :-)

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How is $G\to H$ canonical? Are $G, H$ finitely pro-p generated? –  Bruno Joyal Mar 19 at 13:04
    
@BrunoJoyal Well, $G$ is a free pro-$p$-product over possibly infinitely many inertia groups and $H$ is a free pro-$p$-Galois group. I think neither of them is finitely generated. $G \to H$ is a canonical homomorphism between those groups, but I don't think it plays a role here. Am I correct about this assumption? –  BIS HD Mar 19 at 13:54

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