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I would like to have a simple proof for the following result:

Let $f=\frac{p}{q}:\mathbb{C}\longrightarrow\mathbb{C}$ be a quotient of polynomials (of course, at some points it may be undefined). There is a natural extension to a map $\bar{f}:\mathbb{S}^2\longrightarrow\mathbb{S}^2$, considering $\mathbb{S}^2=\mathbb{C}\cup\infty$ the compactification of the complex numbers (by taking limits, perhaps some analysis is needed). Then $\text{deg}(\bar{f})$(the topological degree of maps between spheres)$=\max\{\text{deg}(p),\text{deg}(q)\}$.

Any idea is welcome.

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    $\begingroup$ If you mean degree as a topological map, the correct formula is $\deg(\bar{f}) = \max(\deg(p), \deg(q))$. $\endgroup$
    – David E Speyer
    Feb 27, 2015 at 16:51
  • $\begingroup$ @DavidSpeyer But what if $p(z)=q(z)=z$? The common degree is 1, but $f$ is constant and its degree is zero. $\endgroup$
    – Jjm
    Feb 27, 2015 at 16:55

1 Answer 1

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The fundamental theorem of algebra tells us that the number of solutions to $f(z) = a$ is the maximum of the degrees of the numerator and denominator.

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