Degree of square map in $\mathbb{P}_{\mathbb{R}}^3$ How can I compute the degree of the map
$f\colon \mathbb{P}_{\mathbb{R}}^3 \rightarrow \mathbb{P}_{\mathbb{R}}^3$
given by $f([x_0:x_1:x_2:x_3])=[x_0^2: x_1^2: x_2^2 :x_3^2]$?
Clearly, a general point in the image has $8$ pre-images and the map is a local diffeomorphism. Since $\mathbb{P}_{\mathbb{R}}^3$ is oriented and connected, then the local cohomologies at those points is the same, so, we have degree $8$ or $-8$. Which one is the degree?
 A: I think that the original count of $4$ pre-images was OK, since a distribution of signs and the opposite distribution of signs represent the same equivalence class. Still this does not mean that the degree has to be $\pm 4$, since you have to count pre-images including a sign. 
Indeed, the degree of $f$ is zero, since $f$ is not surjective. For example $[-1:1:1:1]$ is not in the image. 
A: $\newcommand{\Reals}{\mathbf{R}}$The map $f$ is not a local diffeomorphism. In each affine chart, $f$ is locally represented by the $8$-to-$1$ map
$$
(u_{0}, u_{1}, u_{2}) \mapsto (u_{0}^{2}, u_{1}^{2}, u_{2}^{2}).
\tag{1}
$$
Consequently, $f$ has critical points on the union of the coordinate hyperplanes (i.e., the set of points where some homogeneous coordinate vanishes), since at each such point the differential has rank less than three.
The map (1) (and thus $f$ itself) is rather badly non-surjective. Qualitatively, $f$ acts on the closed upper hemisphere in $S^{3}$ by folding it in half three times; the image of $f$ is the set of lines in $\Reals^{4}$ hitting the non-negative orthant.
