= Solution
Write $w=u+iv$ and $R=1+u^2+v^2$. Inverting the specified <stereographic projection> gives
$$
\phi=\frac1R(2u,2v,1-u^2-v^2).
$$
The coordinate takes the value infinity at the south pole, so a general map is described using the extended complex plane and a second chart there. A pole of $w$ is not necessarily a singularity of $\phi$.
Differentiate the inverse formulas:
$$
d\phi_1=\frac{2du}{R}-\frac{2u\,dR}{R^2},\quad
d\phi_2=\frac{2dv}{R}-\frac{2v\,dR}{R^2},\quad
d\phi_3=-\frac{2dR}{R^2},\qquad dR=2u\,du+2v\,dv.
$$
On squaring and summing, the terms containing $(dR)^2$ cancel, leaving $|d\phi|^2=4(du^2+dv^2)/R^2$. Thus the <stereographic energy of the O3 sigma model> is
$$
\boxed{\widetilde V(w)=4\int_{\mathbb R^2}\frac{|\partial_1w|^2+|\partial_2w|^2}{(1+|w|^2)^2}\,d^2x.}
$$
The same derivatives give $\phi\cdot(\partial_u\phi\times\partial_v\phi)=4/R^2$, with positive sign: at $u=v=0$ the ordered tangent vectors are $(2,0,0)$ and $(0,2,0)$ at the north pole. The chain rule therefore gives
$$
\boxed{N=\frac1\pi\int_{\mathbb R^2}\frac{u_{x_1}v_{x_2}-u_{x_2}v_{x_1}}{(1+|w|^2)^2}\,d^2x
=\frac1\pi\int_{\mathbb R^2}\frac{\operatorname{Im}(\overline{w_{x_1}}w_{x_2})}{(1+|w|^2)^2}\,d^2x.}
$$
This is the <topological degree> for the orientation in the question.
Put $z=x_1+ix_2$, $w_z=(w_{x_1}-iw_{x_2})/2$ and $w_{\bar z}=(w_{x_1}+iw_{x_2})/2$. Expanding the squares gives
$$
\widetilde V=8\int\frac{|w_z|^2+|w_{\bar z}|^2}{(1+|w|^2)^2}\,d^2x,\qquad
N=\frac1\pi\int\frac{|w_z|^2-|w_{\bar z}|^2}{(1+|w|^2)^2}\,d^2x.
$$
For $N\geq0$, subtracting the second expression times $8\pi$ gives
$$
\widetilde V-8\pi N=16\int\frac{|w_{\bar z}|^2}{(1+|w|^2)^2}\,d^2x\geq0;
$$
for $N\leq0$, adding $8\pi N$ gives the analogous square with $w_z$. Consequently
$$
\boxed{\widetilde V\geq8\pi|N|.}
$$
The factor is $8\pi$, not $4\pi$, because the printed energy has no prefactor $1/2$. Equality requires a <holomorphic map> for positive degree, an <antiholomorphic map> for negative degree, or a constant map for zero energy.
Explicit examples for every integer degree are
$$
\boxed{w(z)=z^N\ (N>0),\qquad w(z)=\overline z^{\,|N|}\ (N<0),\qquad w(z)=0\ (N=0).}
$$
To verify both degree and finite energy for $w=z^m$, $m>0$, use polar coordinates: $|w_z|^2=m^2r^{2m-2}$ and $w_{\bar z}=0$. Then
$$
N=2m^2\int_0^\infty\frac{r^{2m-1}}{(1+r^{2m})^2}\,dr
=m\int_0^\infty\frac{ds}{(1+s)^2}=m,
$$
where $s=r^{2m}$. Its energy is $8\pi m$. Complex conjugation changes the degree's sign while preserving the energy, and the constant example has degree and energy zero. These <sigma-model lumps> are smooth sphere-valued maps; the coordinate infinity at spatial infinity is regular in the other target chart.
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