Using determinant of Jacobian to prove injectivity

Question

How would you go about to solve this?

I computed Det of the Jacobian at (1,1) and got that it is not equal to 0 so does that mean it is invertible and thus bijective?

Answer 1

Since

1. $\;\bar{v}$ is continuously differentiable in a neighborhood of $(1,1)$.
2. $\;\det(J_{\bar{v}}(1,1)) \ne 0$.

it follows, by the inverse function theorem, that $\bar{v}$ is injective in a neighborhood of the point $(1,1)$.

However, $\bar{v}$ is not injective on $\mathbb{R^2} \setminus \{(0,0)\}$, since, for example $$\bar{v}\left(0,\sqrt{\pi/2}\right)=\bar{v}\left(0,-\sqrt{\pi/2}\right)$$

Answer 2

The Jacobian is invertible, hence $v$ is locally invertible at $(1,1)$. This excludes the fourth and the third answers. Now, you can see that $v$ is not invertible in $\Bbb{R}^2 \setminus \{ 0 \}$ because $$v(1,1) = v(-1,-1)$$ hence the answer is $(i)$.