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.