Find inverse of a continuous map in $\Bbb R^3$

Question

I need to show that the continuous map $H: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ defined by

$$H = \begin{bmatrix} x_{1} \\ x_{2}+x_{1}^{2} \\ x_{3} + \frac{1}{3}x_{1}^{2} \end{bmatrix} $$

has a continuous inverse. Now I am confused as how to approach this. I know in general $A\cdot A^{-1}=I$, but I do not see how I can apply this to this continuous system. Any suggestions would be more than welcome.

Answer 1

You solve the system$$\left\{\begin{array}{l}x_1=y_1\\x_2+x_1^{\,2}=y_2\\x_3+\frac13x_1^{\,2}=y_3.\end{array}\right.$$You will get$$\left\{\begin{array}{l}x_1=y_1\\x_2=-y_1^{\,2}+y_2\\x_3=-\frac13y_1^{\,2}+y_3.\end{array}\right.$$Therefore$$H^{-1}(x_1,x_2,x_3)=\left(x_1,-x_1^{\,2}+x_2,-\frac13x_1^{\,2}+x_3\right).$$