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Homeomorphism

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Poirot

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Feb 15, 2012
250
Find a homeomorphism between Q={(x,y,z):$x^2+y^6+z^{10}=1$} and the unit sphere in R^3
 

Fernando Revilla

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MHB Math Helper
Jan 29, 2012
661
Re: homeomorphism

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

Edit: I meant $f:S^2\to Q$.
 
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Opalg

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Feb 7, 2012
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Re: homeomorphism

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

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Feb 15, 2012
250
Re: homeomorphism

Does f and f^-1 being continuous mean all partial derivatives exist?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: homeomorphism

Does f and f^-1 being continuous mean all partial derivatives exist?
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.
 
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Poirot

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Feb 15, 2012
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Re: homeomorphism

So what is the definition of f being continuous? Actually, being differentiable is a sufficent condition so that will do
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: homeomorphism

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

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

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Feb 15, 2012
250
Re: homeomorphism

ha ha yes well I suppose I was looking for a sufficent but not neccesary condition - then realised that's just what I had.
 

ModusPonens

Well-known member
Jun 26, 2012
45
Re: homeomorphism

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.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Re: homeomorphism

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 $Q$? (Worried)
Of course, I meant $f:S^2\to Q$. Thanks. :)