Continuous Mapping to Topological Product

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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 if and only if $\pr_1 \circ f$ and $\pr_2 \circ f$ are continuous.


Corollary

Let $T = T_1 \times T_2$ be a 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.


General Result

Let $X$ be a topological space.

Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.

Let $\displaystyle Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$.

For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$th projection on $X$:

$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$


Let $f$ be a mapping from $X$ to $Y$.


Then $f$ is continuous if and only if $\pr_i \circ f$ is continuous for all $i \in I$.


Proof

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 $\pr_2 \circ f$.

$\Box$


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:

\(\displaystyle \map {f^{-1} } {U_1 \times U_2}\) \(=\) \(\displaystyle \map {f^{-1} } {\paren{U_1 \times T_2} \cap \paren{T_1 \times U_2} }\) Cartesian Product of Intersections
\(\displaystyle \) \(=\) \(\displaystyle \map {f^{-1} } {U_1 \times T_2} \cap \map {f^{-1} } {T_1 \times U_2}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {f^{-1} } {\map {\pr_1^{-1} } {U_1} } \cap \map {f^{-1} } {\map {\pr_2^{-1} } {U_2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\paren{\pr_1 \circ f}^{-1} } {U_1} \cap \map {\paren{\pr_2 \circ f}^{-1} } {U_2}\)

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.

$\blacksquare$


Sources