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.
$\blacksquare$