Combination Theorem for Continuous Mappings/Topological Semigroup/Product Rule

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 Set 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.