Does projectivizing always fix problems at infinity? (Or, am I making a mistake somewhere?) This question is motivated by the following homework problem.  I'm trying to explicitly compute the homeomorphism $f:S^2 \rightarrow \mathbb{CP}^1$ by using stereographic projection and considering $\mathbb{CP}^1 = \mathbb{C}\cup {\infty}$.  I'll want to prove that this is an isometry, where $S^2$ has the standard angle metric and $\mathbb{CP}^1$ has the Fubini-Study metric given by $d(\overline{x},\overline{y})=2\cos^{-1}|(x,y)|$, where $x,y\in \mathbb{C}^2$ are unit vectors (and presumably $(-,-)$ is the usual Hermitian inner product).  Later, I'll use this to explicitly compute the Lie group homomorphism $U(2)\rightarrow SO(3)$.
My stereographic projection is from the north pole, takes the equator to the unit circle, and puts the south pole at the origin.  What I've gotten so far is that for $z\not= 1$, \begin{equation*} f(x,y,z)=\left( \frac{x}{1-z} , \frac{y}{1-z} \right) = \frac{x+iy}{1-z} = [x+iy : 1-z ], \end{equation*} where these are coordinates in $\mathbb{R}^2$, $\mathbb{C}$, and $\mathbb{C}\subseteq \mathbb{CP}^1$ respectively.  This is troublesome, because philosophically I'd expect that I should be able to define this for $(x,y,z)\not= (0,0,1)$ and then end up with a function to projective space that extends continuously over the north pole; that's sort of the point of projective space, to make $\infty$ into just another point.  However, it is not immediately obvious that this works, although luckily \begin{equation*} \left| \frac{x+iy}{1-z} \right| = \sqrt{ \frac{|x+iy|^2}{(1-z)^2} } = \sqrt{ \frac{1-z^2}{(1-z)^2}}, \end{equation*} and the limit of this expression as $z\rightarrow 1^-$ is indeed $\infty$.
So, fair enough.  This ends up extending to a continuous function after all.  But: Am I wrong in my philosophical understanding of projective space?
(For what it's worth, I tried using my calculations to verify that $f$ is an isometry, and it didn't look like it was going to work out.  So maybe I really am just doing something wrong.)
 A: One complication in your situation is that you are mixing real and complex coordinates.
If you were considering a map from a complex curve to $\mathbb{CP}^1$, then the kind of computation you are trying to make would work out more straightforwardly.
Because you are looking at a map of real analytic manifolds, not complex analytic ones
(concretely, you are working with the variables $x,y,z$, which are real coordinates), the
point of view you have adopted is perhaps not quite as natural.  Nevertheless, it can be
made to work, as follows:
$$[x+iy:1-z] \text{ (which is where you finished) }
= [x^2 + y^2: (1-z)(x - i y)]$$
$$ = [ 1 - z^2: (1-z)(x-iy)] = [1+z:(x-iy)].$$
This rewriting of your map to $\mathbb{CP}^1$ is now well-defined in a neighbourhood of $(0,0,1)$ on the sphere.  (The fact that I introduced a complex conjugate of $x + i y$ to
facilitate the computation is related to the real vs. complex issue mentioned above.  This is also essentially the same computation you made to check that your map tends to $\infty$ as $z \to 1$, just rewritten in homogeneous coordinates.)
A: "Am I wrong in my philosophical understanding of projective space?"
I think your understanding is incomplete in some ways.
E.g. your remark "resolve singularities by replacing a point with a copy of projective space" doesn't make much sense to me.  
Edit: I think I didn't understand the question correctly. But I couldn't ask any clarification in comments to the question like others do, because of lack of reputation!
It may help in your understanding of "projective spaces" to realize that their resolvement of $\infty$ singularities uses homogene coordinates, e.g. in $\mathbb{CP}^1 = \mathbb{C}\cup {\infty}$ one uses coordiantes $\rho\cdot (z, 1)$ for elements in $\mathbb{C}$ and $\rho (1, 0)$ for ${\infty}$. Then the mapping should involve both coordinates and account for the free factor $\rho \neq 0$.
