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Let $f\colon \mathbb R^2 \to \mathbb R^3$ be defined by the formula $$ f(x,y)=(\sin x,e^y\cos x,xy). $$

Simultaneously $y \geq 0$ and $0 < x < 2\pi$.

The question is whether $f$ is differentiable manifold or not? And why?

Siminore
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user114119
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    It's a bit of an abuse of notation to refer to the function $f$ as a differentiable manifold. I think the question you should be asking is whether the graph of $f$ is a differentiable manifold. – Nick Jan 09 '14 at 17:40
  • The graph or the image? –  Jan 09 '14 at 17:42
  • I do believe I meant the graph, i.e. the subset of $\mathbb{R}^5$ consisting of points $(x,y, \sin x, (e^y)\cos x, xy)$ and subject to the given constraints on $x,y$. – Nick Jan 09 '14 at 17:43
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    $f$ is a parameterization of a surface: http://www.wolframalpha.com/input/?i=plot3d%28sin+x%2C+e%5Ey+cos+x%2C+xy%29 and the response is yes, 'cuz the jacobian $Jf$ has rank two everywhere – janmarqz Jan 09 '14 at 17:44
  • It is a manifold with boundary. – Moishe Kohan Jan 09 '14 at 18:43

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Hint: The inverse function theorem and implicit function theorem allow you to construct a differentiable atlas for the graph of any (nice) smooth function $f: \mathbb{R}^n \to \mathbb{R}^m$. Can you see how?

Nick
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