Are these requirements for a projection between points on spherical surface and scalars possiple? I apologize in advance for my bad language and my usage of incorrect terms.
I need to find a transformation $L: \mathcal{V} \rightarrow \mathcal{S}$ between a scalar value between 0 and 1: $\mathcal{V}=\{x \in \mathcal{R} | 0 \leq x\leq 1\}$ and the set of points on a spherical surface $\mathcal{S}$ defined by a longitude and latitude $(\theta,\phi)$; I do however have a number of specific requirements, which I do not know if make this transformation impossible, the requirements are as follows:


*

*The transformation must be invertible

*The transformation must be continuous (I do not know if this is the correct English word to use, what I mean is that if that is if  $dx\in \mathcal{R}$ and $x, x+dx \in \mathcal{V}$ then $\lim_{dx\rightarrow 0} L(x+dx)=L(x)$). This is not required at the endpoints in $\mathcal{V}$, and the inverse is not required to be (and as far as I can tell can't be) continuous.

*The transformation must "preserve area", that is if two different subsets (properly wrong English word) of the spherical surface $\mathcal{S_1}$ and $\mathcal{S_2}$ have area $A_1$ and $A_2$, and if $l_1$ and $l_2$ is the total length of $L^{-1}(\mathcal{S_1})$ and $L^{-1}(\mathcal{S_2})$, then $l_1/l_2 = A_1/A_2$
I do strongly suspect that these requirements are not possible to achieve (as this seems similar to transforming a spherical surface to a 2D-map, which cannot be done with to strict requirements) my question is, therefore: Can any transformation with these properties exist and if no, why not?
 A: No it is not possible. The requirements 1 and 2 are incompatible.
Notice that $ \ \mathcal{V} \setminus \{ 0,1 \} = \, ] 0,1 [ \ $ is the unit open interval. Let $ \ C = \mathcal{S} \setminus \{ L(0) , L(1) \} \ $. By 1. $ \ L(0) \neq L(1) \ $ and then $C$ is diffeomorphic to a bounded open cylinder. You can see it as a smooth $2$-submanifold (surface) without boundary of $\mathbb{R}^3$. By 1. and 2. altogether $ \ L|_{]0,1[} : \ ]0,1[ \ \to C \ $ is a continuous bijection onto its image $ \ im \! \left( L|_{]0,1[} \right) = L \big[ ]0,1[ \big] = C \ $, ie, $ \ L|_{]0,1[} : \ ]0,1[ \ \to \mathcal{S} \ $ is a continuous injection, but it cannot exist such a function.
Suppose that $ \ L|_{]0,1[} : \ ]0,1[ \ \to C \ $ is a continuous bijection. For all $ \ n \in \mathbb{N}_{\geq 3} = \mathbb{N} \setminus \{ 0,1,2 \} \ $, let $ \ J_n = \left[ \frac{1}{n} , 1 - \frac{1}{n} \right] \ $ and $ \ L_n = L|_{J_n} : J_n \to L[J_n] \subset C \ $. So, for each $ \ n \in \mathbb{N}_{\geq 3} \ $, $J_n \subset \ ]0,1[ \ $, $J_n \, $ is compact and $L_n \, $ is a continuous bijection onto its image $ \ I_n = im(L_n) = L[J_n] \ $, with $ \ I_n \subset C \subset \mathcal{S} \ $. In addition, $ \ I_n = im(L_n) = L[J_n] = \left( L|_{]0,1[} \right) \! [J_n] \ $ is compact in $\mathcal{S} \, $, $\forall n \in \mathbb{N}_{\geq 3} \ $.
Let $ \ n \in \mathbb{N}_{\geq 3} \ $. Since $ J_n \, $ is compact and $\mathcal{S} \, $ is Hausdorff, then $L_n$ is a homeomorphism onto its image $I_n \ $. Again since $ \, I_n \, $ is compact and $\mathcal{S} \, $ is Hausdorff, then  $ \, I_n \, $ is closed in $\mathcal{S} \, $. Hence the closure of $I_n \, $ in $\mathcal{S} \, $ is $ \ \mathcal{C} \ell_{\mathcal{S}} (I_n) = I_n \cong J_n \ $ and it has empty interior. Indeed, if $ \ \mathcal{C} \ell_{\mathcal{S}} (I_n) = I_n \ $ had non-empty interior, it would contain an open ball (relative to the induced metric) and then it would have infinite cut-points, but it cannot be that way, since $I_n \, $ is homeomorphic to $J_n \, $ and $J_n \, $ has just two cut-points, $\frac{1}{n} \, $ and $1 - \frac{1}{n} \ $.
Therefore $ \ C = im \! \left( L|_{]0,1[} \right) = L \big[ ]0,1[ \big] = L \left[ \bigcup_{n \geq 3} J_n \right] = \bigcup_{n \geq 3} L [ J_n ] = \bigcup_{n \geq 3} I_n \ $ is a countable union of nowhere-dense sets. Now, because $ \mathbb{R}^3$ is a complete metric space (with euclidean metric) and because $\mathcal{S} \, $ is a closed subset of $\mathbb{R}^3$, so $\mathcal{S} \, $ is a complete metric space (with the induced metric). By the Baire category theorem, $\mathcal{S} \, $ is a Baire space. Thus every countable union of nowhere-dense sets of $\mathcal{S} \, $ has empty interior. It follows that $ \ C = \bigcup_{n \geq 3} I_n \ $ has empty interior, which is absurd.
Therefore, there is no continuous bijection of the kind $ \, \ ]0,1[ \, \to C \ $.

On the other hand we can have continuous surjections of the kind $ \ [0,1] \to \mathcal{S} \ $, but these are not injections. These functions are called "space-filling curves". In general the Hahn–Mazurkiewicz theorem is the following characterization of spaces that are the continuous image of curves:

A non-empty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected and second-countable space.

