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