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{\bf Question} %note reference to textbook%

Prove, using the definition, that the function $f: \overline{\bf
C} \rightarrow \overline{\bf C}$ given by $f(z) = \frac{2}{z^3}$
for $z\in {\bf C} -\{ 0\}$, $f(0) = \infty$, and $f(\infty) = 0$,
is continuous.

\medskip
Determine whether or not this map $f$ is a homeomorphism of
$\overline{\bf C}$.

\medskip

{\bf Answer}

Note that $f(z)$ is the composition of $J(z)=\ds\frac{1}{z}\
(J(0)=\infty, J(\infty)=0)$ and the polynomial
$g(z)=\ds\frac{1}{2}z^3$ ($g(\infty)=\infty$).

$J$ is continuous by proposition 1.9 (in the book). $g$ is
continuous by exercise 1.14 (in the book), and composition of
continuous functions is continuous (from class). So $f$ is
continuous.

$f$ is \underline{not} a homeomorphism, as it is not injective:

$f(z)=-1$ has 3 solutions, namely $-\sqrt[3] 2,\ -\sqrt[3]
2\omega$, and $-\sqrt[3] 2\omega^2$ where

$\omega=\rm{exp}\left(\ds\frac{2\pi i}{3}\right)$.

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