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Consider the map $w \rightarrow (w^3, w^2)$ from $\mathbb{C}$ to $ \mathbb{C}^2$. Is the image of this map a topological manifold?

I think the map is bijective and continuous, but is not a homeomorphism at $0$. So I don't know whether the image is homeomorphic to $\mathbb{R}^2$ around $(0,0)$.

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Why do you think the map is not a homeomorphism at $0$? –  Thomas Andrews Jan 1 at 22:13
    
The inverse is $y_1 / y_2$ and it's undefined at (0,0) –  Wwk Jan 1 at 22:19
    
But it can be defined at $(0,0)$ so that it is continuous on the image of your map, can't it? –  Thomas Andrews Jan 1 at 22:54
    
The inverse is continuous, but not differentiable, right? –  Wwk Jan 1 at 23:07

1 Answer 1

Your map $f:\mathbb C\to \mathbb C^2: w\mapsto (x,y)=(w^3,w^2)$ is a homeomorphism onto its image (endowed with the subspace topology inherited from $\mathbb C^2$).
Thus the image is a topological manifold since $\mathbb C$ is one.

Since $f$ is clearly continuous and bijective, the only non-trivial point in the argument above is that $f$ is a closed map.
And the reason for that is that $f$ is proper ( proper maps are closed !): indeed, when $w$ tends to infinity $f(w)$ tends to infinty too.

[As an aside, the image of $f$ is the algebraic subset of $\mathbb C^2$ with equation $x^2-y^3=0$ ]

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It's not clear why you introduce $(x,y)$, since you never use that notation. –  Thomas Andrews Jan 1 at 23:39
    
But I do use that notation, my dear @Thomas: where did you stop reading my answer ? –  Georges Elencwajg Jan 1 at 23:41

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