Continuous Mapping to Product Space

Theorem
Let $T = T_1 \times T_2$ be a product space of two topological spaces $T_1$ and $T_2$.

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.

Let $T'$ be a topological space.

Let $f: T' \to T$ be a mapping.

Then $f$ is continuous $\pr_1 \circ f$ and $\pr_2 \circ f$ are continuous.

Necessary Condition
Let $f$ be continuous.

Then by:
 * Projection from Product Topology is Continuous
 * Composite of Continuous Mappings is Continuous

so are $\pr_1 \circ f$ and $\pr_2 \circ f$.

Sufficient Condition
Suppose $\pr_1 \circ f$ and $\pr_2 \circ f$ are continuous.

Let $U_1 \subseteq T_1$ and $U_2 \subseteq T_2$.

Then we have:

Let $U_1$ and $U_2$ be open in $T_1$ and $T_2$ respectively.

Then by continuity of $\pr_1 \circ f$ and $\pr_2 \circ f$:
 * $\map {\paren{\pr_1 \circ f}^{-1} } {U_1}$ and $\map {\paren{\pr_2 \circ f}^{-1} } {U_2}$ are open in $T'$.

So $\map {f^{-1} } {U_1 \times U_2}$ is open in $T'$.

From Natural Basis of Tychonoff Topology of Finite Product, a basis for the product space $T$ is:
 * $\BB = \set{U_1 \times U_2 : U_1 \text{ is open in } T_1, U_2 \text{ is open in } T_2}$

Then we have shown that:
 * $\forall B \in \BB : \map {f^{-1}} B$ is open in $T'$

It follows from Continuity Test using Basis that $f$ is continuous.