Is a map $f\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ invertible iff its induced map on the torus $F\colon[0,1]^{2}/\sim\to[0,1]^{2}/\sim$ is invertible?

Question

Suppose that we have a map $f=(f_{1},f_{2})\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ ($\sim$ stands for the identification of opposite boundaries). Then $f$ induces a map $F\colon[0,1]^{2}/\sim\to[0,1]^{2}/\sim$ on the torus by $$F(x,y):=(f_{1}(x,y) \ \text{mod} \ 1,f_{2}(x,y) \ \text{mod} \ 1),\qquad x,y\in[0,1].$$ Here $\sim$ stands for the identification of opposite boundaries. Is $f$ invertible if and only if $F$ is invertible? If 'iff' is not true, can 'iff' be replaced by a single implication?

EDIT: Maybe I should replace the space $[0,1]/\sim$ by the square $[0,1)^{2}$. Then everything should be well-defined.

No. If $f$ rolls the plane along horizontal lines then glues the ends, then $F$ is the identity but $f$ is far from one-to-one. — Randall, Nov 22 '19 at 14:23
Personally, I think you have not thought through what $x$ and $y$ stand for in your definition of $F(x,y)$. — , Nov 22 '19 at 14:24
How is $F$ well-defined if you mean $F : \mathbb R^2 / \sim \to \mathbb R^2 / \sim$? — , Nov 22 '19 at 14:26
However, by all means there are functions $f:[0,1]\times [0,1]\to \Bbb R^2$ which are injective, but such that $\pi\circ \left. f\right\rvert_{(0,1)\times (0,1)}$ is not injective ($\pi$ being the projection $\pi:\Bbb R^2\to\Bbb R^2/\Bbb Z^2$). For instance, $f(x)=15x$. — , Nov 22 '19 at 14:28
Now it is obvious that $F$ has nothing to do with the behaviour of $f$ outside the square $[0,1]\times [0,1]$. — , Nov 22 '19 at 14:32
Note that $F$ is not usually well-defined if $f_1,f_2$ are arbitrary. But even if it's well-defined, there's no reason to believe that the equivalence holds. However, by covering space theory, if $f$ actually descends to quotients, you can do some stuff — Maxime Ramzi, Nov 22 '19 at 14:33
@Max So what if I replace the space $[0,1]^{2}/\sim$ by the interval $[0,1)$. Then it is always well-defined I think. In this case, can there something be said about invertibility? — Calculix, Nov 22 '19 at 14:39
No, if you do it that way, as pointed out above, $F$ doesn't know anything about $f$ outside of $[0,1)^2$ — Maxime Ramzi, Nov 22 '19 at 14:43

Is a map $f\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ invertible iff its induced map on the torus $F\colon[0,1]^{2}/\sim\to[0,1]^{2}/\sim$ is invertible?

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