TITLE: Projection is an open map
QUESTION [58 upvotes]: Let $X$ and $Y$ be (any) topological spaces. Show that the projection
$\pi_1$ : $X\times Y\to X$ 
is an open map.

REPLY [5 votes]: The projection map
$\pi$ : $X\times Y\to X$ is defined by $\pi(A×B)=A$.
Suppose $U$ is open in $X×Y$. We need to prove that $\pi (U)$ is open in $X$. Since $U$ is open in $X×Y$ there exist basis element $A$ containing a point $x\in X$ and $B$ containing a point $y\in Y$ in $X$ and $Y$ respectively. Therefore $$x\in A=\pi(A×B)\subset\pi (U)$$
Thus $\pi U$ is open in $X$.
Q.E.D.