# Combination Theorem for Continuous Mappings/Topological Semigroup/Product Rule

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## Theorem

Let $\struct {S, \tau_{_S} }$ be a topological space.

Let $\struct {G, *, \tau_{_G} }$ be a topological semigroup.

Let $f, g: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ be continuous mappings.

Let $f * g: S \to G$ be the mapping defined by:

- $\forall x \in S: \map {\paren {f * g} } x = \map f x * \map g x$

Then:

- $f * g: \struct{S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping.

## Proof

Let $f \times g: S \to G \times G$ be the mapping defined by:

- $\forall x \in S : \map {\paren {f \times g} } x = \tuple {\map f x, \map g x}$

From Pointwise Operation is Composite of Operation with Mapping to Cartesian Product:

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

Let $\tau$ be the product topology on $G \times G$.

From Corollary to Continuous Mapping to Topological Product:

- $f \times g : \struct{S, \tau_{_S}} \to \struct {G \times G, \tau}$ is a continuous mapping.

Now $* : \struct {G \times G, \tau} \to \struct {G, \tau_{_G} }$ is continuous by definition of a topological semigroup.

From Composite of Continuous Mappings is Continuous:

- $* \circ \paren {f \times g}$ is continuous.

The result follows.

$\blacksquare$