= Solution
For a smooth map $\phi:M_1\to M_2$ between connected oriented closed manifolds of equal dimension and a volume form $\omega$ on $M_2$, the <topological degree> is defined by
$$
\boxed{\deg\phi=\frac{\int_{M_1}\phi^*\omega}{\int_{M_2}\omega}}.
$$
The standard area form on the unit sphere is
$$
\omega=\frac12\epsilon_{abc}\phi_a\,d\phi_b\wedge d\phi_c.
$$
Since $\int_{S^2}\omega=\operatorname{vol}(S^2)$, this gives
$$
\boxed{\deg\phi=\frac1{2\operatorname{vol}(S^2)}
\int_{S^2}\epsilon_{abc}\phi_a\,d\phi_b\wedge d\phi_c}.
$$
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