Projection from Product Topology is Continuous

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.

Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.

Then both $\pr_1$ and $\pr_2$ are are continuous.

Proof
From Natural Basis of Tychonoff Topology: Finite Product, a basis for $\tau$ is:
 * $\BB = \set {U \times V: U \in \tau_1, V \in \tau_2}$

et $U$ be open in $T_1$.

Then $\map {\pr_1^{-1} } U = U \times T_2$ is one of the open sets in the basis in the definition of product topology.

Thus $\pr_1$ is continuous.

The same argument can be applied to $\pr_2$.

Also see

 * Projection from Product Topology is Open