# 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.

$c_\lambda * f$ and $f * c_\lambda$ are continuous.

Now:

 $\ds \forall x \in S: \,$ $\ds \map {\paren {c_\lambda * f} } x$ $=$ $\ds \map {c_\lambda} x * \map f x$ Definition of $c_\lambda * f$ $\ds$ $=$ $\ds \lambda * \map f x$ Definition of $c_\lambda$ $\ds$ $=$ $\ds \map {\paren {\lambda * f} } x$ Definition of $\lambda * f$

From Equality of Mappings:

$c_\lambda * f = \lambda * f$

Similarly:

$f * c_\lambda = f * \lambda$

The result follows.

$\blacksquare$