Combination Theorem for Continuous Mappings/Topological Semigroup/Multiple Rule

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

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

Let $\lambda \in G$.

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

Let $\lambda * f : S \to G$ be the mapping defined by:
 * $\forall x \in S: \map {\paren{\lambda * f}} x = \lambda * \map f x$

Let $f * \lambda : S \to G$ be the mapping defined by:
 * $\forall x \in S: \map {\paren{f *\lambda}} x = \map f x * \lambda$

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

Proof
Let $c_\lambda : S \to G$ be the constant mapping defined by:
 * $\forall x \in S: \map {c_\lambda} x = \lambda$

From Constant Mapping is Continuous, $c_\lambda$ is continuous.

From Product Rule for Continuous Mappings to Topological Semigroup, $c_\lambda * f$ and $f * c_\lambda$ are continuous.

Now

From Equality of Mappings, $c_\lambda * f = \lambda * f$.

Similarly, $f * c_\lambda = f * \lambda$.

The result follows.