I want to know why if f is an unramified covering to a simply connected surface then this function is an isomorphism?
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1Do you know the relationship between covering spaces and the fundamental group? – Michael Albanese Feb 26 '21 at 16:12
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@Michael Albanese Yes,their deck transformation isomorphic to fundamental group. – Mathgreek Feb 26 '21 at 19:45
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Not quite. Coverings $X' \to X$ up to isomorphism correspond to conjugacy classes of subgroups of $\pi_1(X)$. If $X$ is simply connected, the only cover is $\operatorname{id}_X : X \to X$. – Michael Albanese Feb 26 '21 at 22:01
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@Michael Albanese oh yes, with this fact we could conclude f is an isomorphism,but do we use unramify assumption directly? – Mathgreek Feb 27 '21 at 04:57
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Yes. In my comment above, when I say coverings, I mean unramified coverings, it's not true otherwise. For example, there is a ramified covering $T^2 \to S^2$ despite the fact that $S^2$ is simply connected. – Michael Albanese Feb 27 '21 at 11:20
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Thank you very much – Mathgreek Feb 28 '21 at 08:17