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.

Proof
If $$f$$ is continuous then so are $$\operatorname{pr}_1 \circ f$$ and $$\operatorname{pr}_2 \circ f$$ by Projection from Product Topology is Continuous and Continuity of Composite Mapping.

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.