# Combination Theorem for Continuous Mappings/Topological Semigroup/Multiple Rule

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## 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:

\(\ds \forall x \in S: \, \) | \(\ds \map {\paren {c_\lambda * f} } x\) | \(=\) | \(\ds \map {c_\lambda} x * \map f x\) | 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.

$\blacksquare$