Projection from Product Topology is Open

Theorem

Let $T_1 = \left({A_1, \vartheta_1}\right)$ and $T_2 = \left({A_2, \vartheta_2}\right)$ 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.

General Result

Let $\left \langle X_i \right \rangle_{i \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.

Let $\displaystyle X = \prod_{i \in I} X_i$ be the corresponding product space.

Let $\operatorname{pr}_i : X \to X_i$ be the corresponding projection from $X$ onto $X_i$.

Then $\operatorname{pr}_i$ is open for all $i \in I$.

Proof

If $U$ is open in $T$ then $\operatorname{pr}_1 \left({U}\right) = U_1$ is one of the open sets in $T_1$ by definition.

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

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

$\blacksquare$

Proof of General Result

Let $i \in I$.

Let $U \subseteq X$ be an open set.

Then by the definition of the topology of a product space, $\operatorname{pr}_i \left({U}\right)$ is an open set in $X_i$.

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

$\blacksquare$