Projection from Product Topology is Open and 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 Tychonoff 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 open and continuous.

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

If $U$ is open in $T_1$ then $\map {\pr_1^{-1}} U = U \times T_2$ is one of the sets in the basis $\BB$.

Therefore $\map {\pr_1^{-1}} U$ is open in $\tau$.

Thus $\pr_1$ is continuous.

The same argument applies to $\pr_2$.

Projections are Open
If $U \in \tau$, it follows from the definition of Tychonoff First, we prove that $\operatorname{pr}_1$ is [[Definition:Open Mapping|open.

If $U \in \tau$, it follows from the definition of Tychonoff topology that $U$ can be expressed as:


 * $\displaystyle U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1}} { U_{k,j} }$

where $J$ is an arbitrary index set, $n_j \in \N$, $i_{k,j} \in \set {1,2}$, and $U_{k,j} \in \tau_{i_{k,j} }$.

For all $i \in \set{1, 2}$, define $V_{i, k, j} \in \tau_{i}$ by $V_{i, k, j} = U_{k, j}$ if $i = i_{k, j}$, and $V_{i, k, j} = S_i$ if $i \ne i_{k, j}$.

By definition of projection:
 * $\displaystyle \map {\pr_{i_{k, j} }^{-1}} { U_{k, j} } = \prod_{i \mathop = 1}^2 V_{i, k, j}$

Then:

As $\displaystyle \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1,k,j} \in \tau_1$, it follows that $\pr_1$ is open.

The proof for $\pr_2$ is similsr.