Combination Theorem for Continuous Mappings/Topological Group/Product Rule

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

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

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
By definition, a topological group is a topological semigroup.

Hence $\struct {G, *, \tau_{_G} }$ is a topological semigroup.

From Product Rule for Continuous Mappings to Topological Semigroup:
 * $f * g: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping.

Also see

 * Product Rule for Continuous Mappings to Topological Semigroup