# Homeomorphism

#### Poirot

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

#### Fernando Revilla

##### Well-known member
MHB Math Helper
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

##### MHB Oldtimer
Staff member
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 $Q$?

#### Poirot

##### Banned
Re: homeomorphism

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

#### Opalg

##### MHB Oldtimer
Staff member
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.

#### Poirot

##### Banned
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
Re: homeomorphism

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

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.

#### Poirot

##### Banned
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
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
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 $Q$?
Of course, I meant $f:S^2\to Q$. Thanks.