= Solution
Write the nonconstant bijective map as $p/q$ with coprime <polynomials> and set $d=\max(\deg p,\deg q)$. For a finite value $w$, its finite preimages solve $p(z)-wq(z)=0$. Except for a possible leading-coefficient cancellation value, this <polynomial> has degree $d$.
Multiple roots can occur only at zeros of $p'q-pq'$, since at a root $q\ne0$ by coprimality. This <derivative> numerator is a nonzero <polynomial> for a nonconstant rational map. There are therefore only finitely many finite critical values to exclude. Choose $w$ outside them and outside the leading cancellation value. The <fundamental theorem of algebra> then gives exactly $d$ distinct finite preimages. Bijectivity forces $d=1$.
Hence
$$
\boxed{f(z)=\frac{az+b}{cz+d_0},\qquad ad_0-bc\ne0,}
$$
which is a <Möbius transformation>. Conversely, the inverse is $(d_0w-b)/(a-cw)$, interpreted at its <pole> and at infinity, proving bijectivity on the <Riemann sphere>. The argument also establishes the generic-preimage interpretation of the <degree of a rational map of the Riemann sphere>.
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