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- Thread starter Poirot
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- Jan 29, 2012

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Have you tried $f:Q\to S^2,\; f(x,y,z)=(x,\sqrt[3]{y},\sqrt[5]{z})$ ?

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- Feb 7, 2012

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Shouldn't that be $f(x,y,z)=(x,y^3,z^5)$? Alternatively, doesn't the above map $f$ go from $S^2$ to $Q$?Have you tried $f:Q\to S^2,\; f(x,y,z)=(x,\sqrt[3]{y},\sqrt[5]{z})$ ?

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- Feb 7, 2012

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Not necessarily – you're looking for a homeomorphism, not a diffeomorphism. In other words, the map and its inverse need to be continuous, but not necessarily differentiable.Does f and f^-1 being continuous mean all partial derivatives exist?

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- Feb 7, 2012

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Aren't you supposed to know that by this stage?So what is the definition of f being continuous?

Correct – if you show that the map and its inverse are differentiable then that will imply continuity.Actually, being differentiable is a sufficent condition so that will do

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- Jun 26, 2012

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How about $f:Q\longrightarrow S^2$ such that $f(x,y,z)=\frac{1}{||(x,y,z)||}(x,y,z)$ ?

$f$ is continuous. It's bijective and an open map, so $f^{-1}$ is continuous.

- Jan 29, 2012

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Of course, I meant $f:S^2\to Q$. Thanks.Shouldn't that be $f(x,y,z)=(x,y^3,z^5)$? Alternatively, doesn't the above map $f$ go from $S^2$ to $Q$?