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I am trying to show that:

If $f: X \to Y$ is a non-constant holomorphic map between compact Riemann surfaces, of degree $1$, then $f$ is an immersion.

I tried proving this by contradiction. If $f$ is not an immersion, then there is some point $x$ such that $df_x: T_x(X) \to T_{f(x)}Y$ is not injective. But as $X$ is 1-dimensional, so is $T_x(X)$, and therefore $df_x$ is the zero map.

Does this imply that $f$ would be constant around $x$ which gives us a contradictory, or am I on the wrong path here?

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    $\begingroup$ Around any point where $df_x = 0$, the map (locally, in charts) has to look like $z \mapsto z^n$ where $n \neq \pm 1$. Prove this. Now use the assumption that it's degree 1 to derive a contradiction. $\endgroup$ – user98602 Oct 21 '15 at 20:43
  • $\begingroup$ Is it enough to say that if $f$ looks locally like $z \mapsto z^n$ then $df$ looks like $nz^{n-1}$? $\endgroup$ – Krijn Oct 21 '15 at 20:48
  • $\begingroup$ That's the converse of what I said. $\endgroup$ – user98602 Oct 21 '15 at 20:52
  • $\begingroup$ @MikeMiller Is there an easy way (not using Sard) to see some point in the image of your chart is a regular value? $\endgroup$ – PVAL-inactive Oct 21 '15 at 20:58
  • $\begingroup$ @PVAL The zeros of $df$ are isolated. Since $X$ is compact, there are only finitely many. If $X$ weren't assumed compact, there still would be at most countably many. Hence there are at most finitely (countably) many non-regular values. $\endgroup$ – Daniel Fischer Oct 21 '15 at 21:01

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