Is this map a homeomorphism onto its image?

Question

Let us consider the map $\phi: \, (-1,\infty) \times \mathbb{R} \to \mathbb{R}^3$ so defined: $$ \phi(x,y)=\left(\frac{3x}{1+x^3},\frac{3x^2}{1+x^3},y\right). $$ Is it a homeomorphism onto its image? I think no, since the map $\phi$ represents a surface in $\mathbb{R}^3$ whose shape is the similar to that of the Descartes' folium. For any $C^\infty$-map $\phi: U \to \mathbb{R}^3$ whose differential has rank 2, it can be showed that it is locally a homeomorphism onto its image. So, I considered an open like $V:=(-1,1) \times (-1,1)$. What can I say of its image? Is it open? If no, I proved that $\phi$ it is not a homeomorphism onto its image, but I can't see which is the image of $V$. Can you help me, please?

Answer 1

The map $\phi$ is not a homeomorphism onto its image $S\subset{\mathbb R}^3$. The surface $S$ is an infinite vertical cylinder that intersects the $(x,y)$-plane in a curve which is part of a folium of Descartes, see the following figure:

This curve does not have a self-intersection, but almost. The inverse map $\phi^{-1}$ is well defined, but is not continuous at the points $\phi(0,y)=(0,0,y)$. Any neighborhood of $(0,0,0)$ contains points of $S$ that are mapped by $\phi^{-1}$ onto points in $(-1,\infty)\times{\mathbb R}$ of the form $(M,0)$ with $M\gg1$, hence lying far away from $\phi^{-1}(0,0,0)=(0,0)$.

Answer 2

For this question you should change your approach. You do not need to calculate the image of $\phi$ to show that $\phi$ is a homoemorphism onto its image. A homeomorphism is a continuous bijectiv map such that the inverse map is continuous as well. In other words you have to show the following:

1. $\phi$ is injectiv. This shows the existence of an inverse map.

2. The rank of the differential of $\phi$ is 2. As you already pointed out, this would show that $\phi$ is a local homeomorphism. This would mean the inverse map is locally continuous, and since continuity is a local property it is globally continuous.

Please note: If the first property fails, $\phi$ is not a homeomorphism. If the second property fails it could still be homeomorphism. If it fails in a point $p$, you have to examine this point $p$. Only at $p$ the continuity of the inverse map can break down.