a semicubical parabola $L$ in $\mathbb C^2$ is given by $y^2=x^3$.
I showed that a bijective function $f\colon\mathbb C \to L$ defined by $t \mapsto (t^2, t^3)$ becomes a homeomorphism regarding the Zariski topolgy.
But it doens't seem clear to me whether $f$ is a homeomorphism with the usual topoology on the complex numbers. Anyone could give me some proof?
In my opinion it is much easier to show the result for the Euclidean topology in a direct way: The map $f$ defined above is continous with regard to the Euclidean topology. Its inverse is the map
$$g: L \longrightarrow \mathbb C, \ g(x,y):= y/x \ for\ (x,y) \neq (0,0) \ and \ g(0,0):= 0.$$
To show that $g$ is continous in $(0,0)\in L$:
$$lim_{x \rightarrow 0} \ (y/x)^2 = lim_{x \rightarrow 0} \ (y^2/x^2) = lim_{x \rightarrow 0} \ x= 0,$$
hence also $lim_{x \rightarrow 0} \ (y/x) = 0$, q.e.d.