# The open mapping theorem

The open mapping theorem is often presented as a corollary of RouchÃ©'s theorem.
However it is possible to prove the open mapping theorem using less theoretical and more elementary methods.
## Open mappings

Let $X$ and $Y$ be topological spaces. A map $f:X \rightarrow Y$ is an
*open mapping* iff for all open sets $U$ in $X$ the set $f(U)$ is open in
$Y$.
## The squaring function in real and complex analysis

As a function of a real variable the squaring function $f(x) = x^2$ is not an
open mapping. For example the image of the open set $(-1, 1)$ under $f$ is $[0,
1)$ which is closed. However as a function of a complex variable $f(z) = z^2$ is
an open mapping. To see this we check that the image of any open ball $U$ in
$\mathbb C$ under $f$ is open. On the one hand if $0 \notin U$ then $f'(z)\neq
0$ and so by the inverse function theorem $f(U)$ is open. On the other hand if
$0\in U$ we can decompose $U$ into $B_0(\epsilon)\cup V$ where $0\notin V$ and
$\epsilon > 0$. But the image $f(B_0(\epsilon)) = B_0(\sqrt{\epsilon})$ which is
open and by our previous argument $f(V)$ is open. Therefore $f(U)$ is open.
## Polynomials in real and complex analysis

We have just seen that not every polynomial is an open mapping when considered
as a function of a real variable. However as a function of a complex variable
any non-constant polynomial $p(z) = \sum_{i=0}^n c_i z^i$ is an open mapping.
Again if $U$ is any open ball centered at $z_0$ where $p'(z_0)\neq 0$ then
$p(U)$ is open by the inverse function theorem. If $p'(z_0) = 0$ then we can
write (using the assumption that $p$ is non-constant) \[p(z) = c_k (z-z_0)^k
h(z)\] where $k\in {\mathbb N}$, $c_k\in {\mathbb C}$ and $h(z_0) = 1$. Since
$h$ is continuous (it is in fact a polynomial) we can choose $\delta > 0$ such
that $z\in B_{z_0}(\delta)$ implies $\|h(z)\| > \frac{1}{2}$. But then the image
$f(B_{z_0}(\delta))$ contains $B_0\left(c_k\delta^{\frac{1}{k}}/2\right)$.
## The open mapping theorem

In fact every function of a complex variable is either constant or an open
mapping. To see this we combine the previous section with Taylor's theorem..