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Consider this definition of the parametrization of a manifold, found in Hubbard & Hubbard:

A parametrization of a $k$-dimensional manifold $M\subset\mathbb{R}^n$ is a mapping $\gamma:U\subset \mathbb{R}^k\to M$ satisfying:

1) $U$ is open.

2) $\gamma$ is $C^1$, and bijective with $M$.

3) $[D\gamma(u)]$ is 1-1 for all $u\in U$.

My question is this: if instead we start with a parametrization and replace $M$ with the image of $\gamma$, will the image in fact be a manifold? In Hubbard this seems to be suggested, but never stated explicitly (as far as I can see).

Maybe the point is that we can locally transform a parametrization into the graph of a $C^1$ function...

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Are manifolds specifically considered to be $C^1$ manifolds in this book? –  Zev Chonoles Feb 7 '13 at 13:46
    
I think so. They must locally be graphs of $C^1$ functions. –  Wouter Zeldenthuis Feb 7 '13 at 13:49
    
When you talk about Hubbard, are you referring to John Hubbard who co-authored this book with his wife? –  mez Feb 7 '13 at 18:08
    
I think you have to clarify what you mean by a manifold $M \subset \Bbb{R}^n$. $M$ is clearly a $C^1$ manifold with the topology induced by $U$, but is not necessarily a $C^1$ submanifold of $\Bbb{R}^n$, as manu-fatto's example below shows. –  Nils Matthes Feb 7 '13 at 22:44
    
As noted below, if you're looking at a $\mathcal{C}^1$ submanifold of $\mathbb{R}^n,$ you also need the inverse function to be $\mathcal{C}^1.$ –  lyj Feb 8 '13 at 4:58

2 Answers 2

No, it is not. Consider for example a figure 8 curve $$ (x(t),y(t)) = (\sin(t),\sin(2t),\qquad t\in (0,2\pi). $$ You should require also the inverse function to be continuous, if you want it to be a manifold.

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My own attempted answer is that this is not true. Consider $$\gamma(s,t)=\left(\frac{t}{1-u},\frac{t^2}{1-u},\frac{t^3}{1-u}\right).$$ Its image is the solution set to $xz=y^2$, minus the x-axis and z-axis (but the origin we will keep).

This set is locally the graph of a function except at the origin, where it is plainly not!

How is this?

EDIT: I changed the parametrization.

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$D\gamma(s,t)$ does not have full rank when $s=t.$ What conditions are you assuming hold for your parametrization? –  lyj Feb 7 '13 at 15:46
    
The conditions in the question statement, which define a parametrization. –  Wouter Zeldenthuis Feb 7 '13 at 15:49
    
Then your example is not a counter-example? –  lyj Feb 7 '13 at 15:54
    
I modified the parametrization. –  Wouter Zeldenthuis Feb 7 '13 at 15:59
    
what is $u$ in the above parametrization? and more importantly, what is the domain of the parametrization? –  Nils Matthes Feb 7 '13 at 21:50

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