Surjectivity of a Three-Dimensional Function with Non-Negative Real Inputs

[atomqwerty]

Hello,

is this function surjective?

\Phi : \Re^{+} \diamond \Re \diamond \Re \rightarrow \Re^{3}<br /> (r,\varphi,\theta) \rightarrow (r cos\varphi sin\theta, r sin\varphi sin\theta, r cos\theta)

PS Diamond means X (cross)

 

[dimitri151]

It is not surjective.

 

[atomqwerty]

It is not surjective.

why?

thanx

 

[dimitri151]

I mean it is surjective.

 

[Petr Mugver]

It is surjective, because for every a = (x, y, z) in R^3 you can find b = (r, theta, phi) mapped to a by your function. Namely,

r = sqrt (x^2 + y^2 + z^2)

theta = arctan (sqrt(x^2 + y^2) / z)

phi = arctan(y / x)

(this is "almost true", in the sense that you have to be careful with the arctan).

 

[atomqwerty]

It is surjective, because for every a = (x, y, z) in R^3 you can find b = (r, theta, phi) mapped to a by your function. Namely,

r = sqrt (x^2 + y^2 + z^2)

theta = arctan (sqrt(x^2 + y^2) / z)

phi = arctan(y / x)

(this is "almost true", in the sense that you have to be careful with the arctan).

Ok, so that implies x != 0 and z!=0 (!= means not equal to), that implies at the same time

rcosT = 0
rcosPsinT=0

(T=theta, P=Phi)

The first leads us to r=0 or T=(2n+1)Pi/2, with n=1,2,3,...
Second leads moreover to P=(2n+1)Pi/2, with n=1,2,3,... (because SinT == +-1)

so we have to redefine the range of the function from [R+ x R x R] to [R+ x (0,Pi/2) x (0,Pi/2)] OR keep the original range but rstricting to [R+ x R- {(2n+1)Pi/2} x R- {(2n+1)Pi/2}]

Is this correct?

thank

 

[mathwonk]

aren't these just the spherical coordinate map? hence clearly surjective by the geometry of the situation. i.e. every point in space does have spherical coordinates.

 

[atomqwerty]

aren't these just the spherical coordinate map? hence clearly surjective by the geometry of the situation. i.e. every point in space does have spherical coordinates.

Yes, it is, but what I'm saying is that we would need to redefine the range RxRxR to Rx[0,2Pi)x[0,Pi) so the coordinates can be one to one.

 

[Petr Mugver]

Yes, it is, but what I'm saying is that we would need to redefine the range RxRxR to Rx[0,2Pi)x[0,Pi) so the coordinates can be one to one.

Well, this was not your initial question. Your map is surjective but not one to one... consider the origin, or the north and south poles for example. In general, you need several "sheets" or "charts" in the language of manifolds.

 

[reyomit]

Forgive me if I am missing something, but is there anything mapping to (0,0,0)?

 

[dimitri151]

r=0.

 

[reyomit]

I was under the impression that \Re^{+} denoted the positive reals and not the non-negative reals.

 

[dimitri151]

For some reason I read that as non-negative reals. If you can't have r=0 then you can't get the right side to equal (0,0,0).

 

[dimitri151]

Z⁺ is the non-negative integers so I read R⁺ as the non-negative reals so r=0 is allowed.