Metric on a Quotient of the Riemann Sphere Let $P$ denote the quotient space obtained by the action of $\mathbb{Z}\backslash2\mathbb{Z}$ by the antipodal map $z\mapsto\frac{1}{z}$ on the riemann sphere $\hat{\mathbb{C}}$ (identified here with $\mathbb{C}\cup\left\{\infty\right\}$). I identify $P$ with the set:
$\left\{ z\in\mathbb{C}:\left|z\right|<1\right\} \cup\left\{ e^{it}:0\leq t\leq\pi\right\} $ 
that is to say, I use elements of the above set as the representatives for the equivalence classes in $P$. I am looking for formula for a function $f:P\times P\rightarrow[0,\infty)$ such that $f$ is a metric on $P$. Specifically, I would like a formula for $f$ that I can evaluate by plugging in representative elements in the above set (or something like that, more or less).
Thanks in advance.
Edit: forgive me for sounding desperate, but I cannot make due with an explanation of how to obtain such a formula. I want the formula. 
An analogy for you if I have yet to make myself clear: suppose I was asking for the area of a square with side length $s$. The answers I have received so far for my question are akin to saying "multiply $s$ by itself" or "use the area formula for a square". The answer I am looking for is akin to saying "$s^{2}$". I need the formula. And please, no expressions with differentials, nor matrices, or any of that. I need to know how to compute the metric by using the complex numbers in the indicated set that I have identified with $P$.
 A: This answer necessarily omits important details because there is not a unique metric on a sphere, because the term metric can refer to different types of structure (particularly, to a "topological metric" or to a Riemannian matric), and because the intended purpose of the requested metric on the quotient has not been explained.

A complex number $w = u + iv$ with $u$ and $v$ real corresponds, under stereographic projection from the north pole, to
$$
(x, y, z) = \frac{(2u, 2v, u^{2} + v^{2} - 1)}{u^{2} + v^{2} + 1}.
\tag{1}
$$
The chordal metric on the sphere is defined by
$$
d((x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}))
= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}}.
\tag{2a}
$$
The great circle metric on the sphere is defined by
$$
d'((x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}))
= \arccos(x_{1}x_{2} + y_{1}y_{2} + z_{1}z_{2}).
\tag{2b}
$$
Combining, you get a metric
$$
d(w_{1}, w_{2}) = d(u_{1} + iv_{1}, u_{2} + iv_{2})
$$
on the complex plane by using (1) to associate points $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ to your complex numbers, then using (2a) or (2b) or some other metric of your choosing on the sphere.
The induced metric on the quotient of the sphere by the involution $w = u + iv \mapsto \frac{1}{w} = \frac{u - iv}{u^{2} + v^{2}}$ can be found by taking two complex numbers $w_{1}$ and $w_{2} \neq w_{1}$, $1/w_{1}$ and using the scheme of the preceding paragraph to find the smaller of $d(w_{1}, w_{2})$ and $d(1/w_{1}, w_{2})$.
This algorithm can, in principle, be expressed as an explicit formula, but to do so would be uselessly messy.
A: FYI what you named $P$ is the projective plane $\mathbb {RP}^2$. It's a really really classical and super-important manifold. (You don't want to forget it!)
By the way, your identification is a bit clumsy. I am not convinced it's even true. Anyway, I don't recommend you to use it.
There are general results about getting a metric on a quotient space given by the action of a group satisfying a few properties. (You should look into that.)
In this present case, let's take $p,q\in P$. By definition, $p$ is class of two elements in $\widehat{\mathbb C}$ (let's say $p_1$ and $p_2=1/p_1$), and so is $q$. Well, you can compare the distance between $p_1$ and $q_1$, $p_2$ and $q_1$, $p_2$ and $q_2$, and, $p_1$ and $q_2$. What can you tell about the minimum value of theses ?
A: The sphere is compact, and it's quotient by the antipodal map is homeomorphic to the projective plane.  The standard metric on the Riemann sphere is addressed in this question:
How to show that the spherical metric satisfies the triangle inequality?
Since any point in the quotient has only two preimages, any pair of points will have four preimage pairs.  Then you can define the distance on the quotient as 
$$d(x,y)=\min \left\{d_{S^1}(\tilde{x},\tilde{y})\mid \tilde{x}\in p^{-1}(x),\tilde{y}\in p^{-1}(y)\right\}$$
It's straightforward to check that this satisfies the triangle inequality.  It's also "explicit" in the sense that the minimum is taken over a finite set of 4 points.  Given $x,y$ in the quotient, it's easy to calculate all the preimages and then use the formula for the distance on the sphere.
Edit:  After looking more closely at the link I provided, I realize it doesn't give an explicit formula for the distance function on the sphere.  Here is an outline of how to get one.
First, we need a formula for the distance between two points on the standard sphere $$S^2=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y+2+z^2=1\}$$
To get this, we take $v_i=(x_i,y_i,z_i)$ for $i=1,2$ and note that the geodesic connecting them lies on a great circle, which will necessarily have radius 1.  The distance between them will be the length of an arc on that great circle.  The length of an arc on a circle of radius $r$ is equal to $r\theta$ where $\theta$ is the angle formed by the radii which determine the arc.  In this case $\theta$ is precisely the angle formed by the two vectors in $\mathbb{R}^3$.  We can compute this angle using the dot product:
$$d(v_1,v_2)=|\cos^{-1}(v_1\cdot v_2)|$$
Next, we need to translate this into a formula on $\mathbb{C}\cup\infty$ using the stereographic projection.  There is an explicit formula for stereographic projection here: Stereographic Projection
You can compute the inverse of this map, then pull back the metric on the sphere to get a formula for the metric on $\mathbb{C}$.  The formula you get will probably still work when one of the points is $\infty$.
A: The map $z \mapsto \frac{1}{z}$ is the non-trivial covering (deck) transformation of the branch covering holomorphic rational map $f : \mathbb{\hat{C}
} \to \mathbb{\hat{C}}$ given by the formula (which you can post-compose with any linear-fractional transformation if you wish) 
$$f(z) = \frac{1}{2} \, \left(z + \frac{1}{z} \right)$$ This a double branch covering from the Riemann sphere to the Riemann sphere. So your quotient space is the sphere (however, strictly speaking, the quotient space is an orbifold, but that is not crucial). So simply take any metric on the round sphere, project it to $\mathbb{C}$ by stereographic projection and you have a metric on your quotient space. Then, if you want to use it on the original Riemann sphere, simply pull it back by the map $f$ that I wrote above.  
