# Continuous Mapping to Product Space/Corollary

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## Corollary to Continuous Mapping to Product Space

Let $T = T_1 \times T_2$ be the product space 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

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.

From Continuous Mapping to Product Space, $f \times g$ is continuous if and only if $\pr_1 \circ \paren {f \times g}$ and $\pr_2 \circ \paren {f \times g}$ are continuous.

Now:

\(\ds \forall x \in T': \, \) | \(\ds \map {\pr_1 \circ \paren {f \times g} } x\) | \(=\) | \(\ds \map {\pr_1} {\map {\paren {f \times g} } x}\) | Definition of Composite Mapping | ||||||||||

\(\ds \) | \(=\) | \(\ds \map {\pr_1} {\tuple {\map f x, \map g x} }\) | Definition of Mapping $f \times g$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map f x\) | Definition of First Projection from $T$ |

From Equality of Mappings, $\pr_1 \circ \paren {f \times g} = f$.

Similarly:

- $\pr_2 \circ \paren {f \times g} = g$

The result follows.

$\blacksquare$