Why does simple counter-example to global inverse theorem not work?

Question

Theorem 2.1 in [1] states that

Let $M$ and $N$ be connected, oriented, d-dimensional $C_1$-manifolds, without boundary. Let $f: M\mapsto N$ be a proper $C^1$-map such that the Jacobian $J(f)$ never vanishes. Then $f$ is surjective. If $N$ is simply connected in addition, then $f$ is also injective.

Here is a simple "counter-example" for which I don't see where it contradicts the theorem: Let $M, N = S^2$ be spheres and let $f([\alpha_1, \alpha_2]) = [2\alpha_1, 2\alpha_2]$ (where $[\alpha_1, \alpha_2]$ are the two angles describing the direction of a unit vector). $M, N$ are simply connected, oriented and differentiable manifolds without boundary. $f$ is smooth, its Jacobian never vanishes (on the manifold) and it is proper because any compact set in $N$ is mapped to the union of two compact sets in the pre-image. But $f$ is not bijective, which seemingly contradicts the theorem?

[1] On global inversion of homogeneous maps, M. Ruzhansky and M. Sugimoto, https://arxiv.org/pdf/1305.5930.pdf

Answer 1

I will assume that OP is using the spherical coordinates

$$ \Phi(\alpha_1,\alpha_2) = (\sin\alpha_2\cos\alpha_1, \sin\alpha_2\sin\alpha_1, \cos\alpha_2), $$

where $\mathbb{S}^2$ is regarded as the unit sphere in $\mathbb{R}^3$. If we write $\tilde{f}(\alpha_1, \alpha_2) = (2\alpha_1, 2\alpha_2)$, then

$$f = \Phi \circ \tilde{f} \circ \Phi^{-1}$$

yields a well-defined function over $\mathbb{S}^2$. However, since $\Phi$ fails to give rise to a coordinate chart near both poles of $\mathbb{S}^2$, it may obscure the true behavior of $f$ near the point $p$ when either $p$ or $f(p)$ are the poles of $\mathbb{S}^2$.

To avoid this issue, we first note that $f$ is written as

$$ f(x, y, z) = \left( \frac{2z(x^2-y^2)}{\sqrt{x^2+y^2}}, \frac{2z(2xy)}{\sqrt{x^2+y^2}}, 2z^2-1 \right) $$

using the Cartesian coordinates $(x,y,z)$ inherited from that of $\mathbb{R}^3$. Together this and appropriate choices of coordinate charts $\varphi$ and $\psi$ around $p$ and $f(p)$ respectively, we can study the behavior of $f$ in a more transparent way.

1. $f$ maps the north pole $p = \mathbf{e}_3$ to itself. So it is convenient to choose both $\varphi$ and $\psi$ as the map from the upper hemisphere $\mathbb{S}^2 \cap \{z>0\}$ to the unit disk $\mathbb{D}=\{(x,y) : x^2+y^2 < 1\}$ via the projection $$(x,y,z)\mapsto(x,y).$$ Then $$ (\psi\circ f \circ \varphi^{-1})(x, y) = 2\sqrt{\frac{1-x^2-y^2}{x^2+y^2}} \left( x^2-y^2, 2xy \right). $$ This map fails to be differentiable at $(0, 0)$.

2. $f$ maps $p = \mathbf{e}_1$ to the south pole. Let $\varphi : \mathbb{S}^2\cap\{x > 0\} \to \mathbb{D} $ and $\psi : \mathbb{S}^2\cap\{z < 0\} \to \mathbb{D} $ as $$ \varphi(x, y, z) = (y, z) \qquad\text{and}\qquad \psi(x, y, z) = (x, y). $$ Then $$ (\psi\circ f \circ \varphi^{-1})(y, z) = \left( \frac{2z(1-2y^2-z^2)}{\sqrt{1-z^2}}, \frac{4yz\sqrt{1-y^2-z^2}}{\sqrt{1-z^2}} \right), $$ and a direct computation tells that its Jacobian determinant at $(y, z) = (0, 0)$ vanisehs.