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I am currently reading "Lectures on Harmonic Maps" by R. Schoen and S. T. Yau and have problems understanding one step in the proof of a Corollary on p. 13.

Corollary: Suppose $N = S^2$, and $M, S^2$ are equipped with arbitrary metrics. If $u: M \to S^2$ is harmonic and $\mathrm{deg}(u) > g_M - 1$, then $u$ is holomorphic.

Corollary: There is no harmonic map from $T^2$ to $S^2$ with $\mathrm{deg}(u) = 1$.
Proof: Suppose $u$ is such a harmonic map. Since $\mathrm{deg}(u) = 1 > \mathrm{genus}(T^2) - 1 = 0$, $u$ is holomorphic by the previous corollary, i.e., $\vert \overline{\partial}u \vert \equiv 0$. Hence $J(u) \geq 0$. Since $u$ has degree $1$, we then see that $\#(u^{-1}(q)) = 1$, for any regular value $q \in S^2$. On the other hand, $u$ is holomorphic, and hence is a branched covering; thus $J(u) > 0$ everywhere and $u$ is a diffeomorphism. This contradiction shows that there is no such $u$.

My question is: How do we know that $J(u) > 0$ everywhere?

chrstphfrtz
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  • Compute the Jacobian matrix of a holomorphic map from a surface to a surface and confirm that its determinant is always positive. – Deane Feb 20 '24 at 16:15
  • They garbled the proof: the sign of the Jacobian is irrelevant for the proof of a local diffeomorphism. What is important is that the map has no critical points since it is 1-1. The latter is a nice exercise in complex analysis: An injective holomorphic function has nowhere vanishing derivative. – Moishe Kohan Feb 20 '24 at 17:00
  • Yes, that is exactly what I don't understand. Why is the map 1-1 everywhere? I just know that for the regular values because I have a degree $1$ map, right? Are all points $q \in S^2$ regular values? – chrstphfrtz Feb 20 '24 at 21:09
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    Remember that a holomorphic map $\Bbb C\to\Bbb C$ looks locally like $f(z)=z^m$ for some nonnegative integer $m$. – Ted Shifrin Feb 20 '24 at 22:41
  • See the 2nd answer at https://math.stackexchange.com/questions/54973/injective-holomorphic-functions-that-are-not-conformal – Moishe Kohan Feb 21 '24 at 01:45
  • I think I see it now: Harmonic maps are conformal in dimension 2 and $u$ is holomorphic, therefore, $J(u) \neq 0$ everywhere and even $J(u)>0$ everywhere since it is a degree 1 map and preserves orientation, correct? This means that due to the Inverse Function Theorem I get a local diffeomorphism in a neighborhood of every point. Due this this question math.stackexchange.com/q/4509942/1172103 this already suffices to establish the contradiction. If the reasoning is correct I can write an answer myself if no one else wants/has time to do so. – chrstphfrtz Feb 21 '24 at 10:57

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