Continuous Mapping to Product Space

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

Let $T'$ be a topological space.

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

Then $f$ is continuous iff $\operatorname{pr}_1 \circ f$ and $\operatorname{pr}_2 \circ f$ are continuous.

Necessary Condition
Let $f$ be continuous.

Then by: so are $\operatorname{pr}_1 \circ f$ and $\operatorname{pr}_2 \circ f$.
 * Projection from Product Topology is Continuous
 * Continuity of Composite Mapping

Sufficient Condition
Suppose $\operatorname{pr}_1 \circ f$ and $\operatorname{pr}_2 \circ f$ are continuous.

Let $U_1 \subseteq T_1$ and $U_2 \subseteq T_2$. Then we have:

So if $U_1$ and $U_2$ are open in $T_1$ and $T_2$, then $\left({\operatorname{pr}_1 \circ f}\right)^{-1} \left({U_1}\right)$ and $\left({\operatorname{pr}_2 \circ f}\right)^{-1} \left({U_2}\right)$ are open in $T'$ by continuity of $\operatorname{pr}_1 \circ f$ and $\operatorname{pr}_2 \circ f$.

So $f^{-1} \left({U_1 \times U_2}\right)$ is open in $T'$.

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