Continuous Mapping to Topological Product

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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 if and only if $\operatorname{pr}_1 \circ f$ and $\operatorname{pr}_2 \circ f$ are continuous.

Corollary

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


Let $T'$ be a topological space.

Let $f: T' \to T_1$ be a mapping.

Let $g: T' \to T_2$ be a mapping.


Let $f \times g : T’ \to T$ be the mapping defined by:

$\forall x \in T’ : \map {\paren {f \times g}} x = \tuple{ \map f x, \map g x}$


Then $f \times g$ is continuous if and only if $f$ and $g$ are continuous.


Proof

Necessary Condition

Let $f$ be continuous.

Then by:

Projection from Product Topology is Continuous
Composite of Continuous Mappings is Continuous

so are $\operatorname{pr}_1 \circ f$ and $\operatorname{pr}_2 \circ f$.

$\Box$


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:

\(\displaystyle f^{-1} \left({U_1 \times U_2}\right)\) \(=\) \(\displaystyle f^{-1} \left({\left({U_1 \times T_2}\right) \cap \left({T_1 \times U_2}\right)}\right)\) Cartesian Product of Intersections
\(\displaystyle \) \(=\) \(\displaystyle f^{-1} \left({U_1 \times T_2}\right) \cap f^{-1} \left({T_1 \times U_2}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle f^{-1} \left({\operatorname{pr}_1^{-1} \left({U_1}\right)}\right) \cap f^{-1} \left({\operatorname{pr}_2^{-1} \left({U_2}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\operatorname{pr}_1 \circ f}\right)^{-1} \left({U_1}\right) \cap \left({\operatorname{pr}_2 \circ f}\right)^{-1} \left({U_2}\right)\)

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

Then by continuity of $\operatorname{pr}_1 \circ f$ and $\operatorname{pr}_2 \circ f$:

$\left({\operatorname{pr}_1 \circ f}\right)^{-1} \left({U_1}\right)$ and $\left({\operatorname{pr}_2 \circ f}\right)^{-1} \left({U_2}\right)$ are open in $T'$.

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

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

$\blacksquare$

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