Show that an open mapping need not map closed sets onto closed sets

Question

My question is the following one

Question - Show that an open mapping need not map closed sets onto closed sets

I have no idea of how to solve this particular problem. Do you have any good suggestions?

Thanks in advance!

Answer 1

One example is the projection $\pi: \mathbb R^2 \rightarrow \mathbb R$ given by $$\pi(x, y) = y.$$ This map is open by definition of the product topology, but it is not closed. To see this, consider the set $$ A = \{(x, y) \in \mathbb R^2 \mid y \leq \text{arctan}(x)\}.$$ Then $A$ is closed, but $\pi(A) = (-\infty, \pi/2)$ is not.

Answer 2

Another is the mapping $f: X\longrightarrow Y$ where $X=Y=\mathbb{R}$, defined by $x\longmapsto e^x$. This mapping is an open mapping because the inverse function obtained by restricting the range $f^{-1}: (0,\infty)\longrightarrow X$ is continuous. However, $X$ is closed in $X$, but $f(X)=(0,\infty)$ is open in $Y$.