Let $\alpha : \mathbb{R}^2 \to \mathbb{R}^3$ defined by $\alpha(x,y) = (x(x^2+y^2), y(x^2 + y^2), x^2 + y^2)$. How do I show that $\alpha$ as a map from $\mathbb{R}^2$ to its image, it has a continuous inverse - ie. $\alpha^{-1}: \alpha(\mathbb{R}^2) \to \mathbb{R}^2$ is continuous?
This was given as an example in Munkres' Analysis on Manifolds, regarding a function that was almost a coordinate patch (it fails to be due to the rank of the matrix $D\alpha (x)$ failing to be of rank 2 at 0), but I can't seem to prove rigourously why the inverse should in fact be continuous.
Yes, your solution is fine. Just to finish this off slightly differently, note that $M$ is a (topological) $2$-dimensional manifold parametrized by $(a,b)$. I.e., $M= \{(a,b,c): c=(a^2+b^2)^{1/3}\}$. Now, for $(a,b,c)\in M$ with $c\ne 0$, $$\alpha^{-1}(a,b,c) = \left(\frac ac,\frac bc\right) = \frac1{(a^2+b^2)^{1/3}}(a,b).$$ Note that $$\left\|\frac{(a,b)}{(a^2+b^2)^{1/3}}\right\| = \frac{(a^2+b^2)^{1/2}}{(a^2+b^2)^{1/3}} = (a^2+b^2)^{1/6},$$ whose limit as $(a,b)\to (0,0)$ is clearly $0$. (Yes, if $\|(a,b)\|<\delta$, then $\|\alpha^{-1}(a,b,c)\|<\delta^{1/3}$, as you said.)