Projection from Product Topology is Open

Theorem
Let $T_1 = \left({S_1, \tau_1}\right)$ and $T_2 = \left({S_2, \\tau_2}\right)$ be topological spaces.

Let $T = T_1 \times T_2$ be the topological product of $T_1$ and $T_2$.

Let $\operatorname{pr}_1: T \to T_1$ and $\operatorname{pr}_2: T \to T_2$ be the first and second projections from $T$ onto its factors.

Then both $\operatorname{pr}_1$ and $\operatorname{pr}_2$ are open.

Proof
If $U$ is open in $T$ then $\operatorname{pr}_1 \left({U}\right)$ is one of the open sets in $T_1$ by definition.

Thus $\operatorname{pr}_1$ is open.

The same argument applies to $\operatorname{pr}_2$.