Combination Theorem for Continuous Mappings/Topological Group/Inverse Rule
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Theorem
Let $\struct {S, \tau_S}$ be a topological space.
Let $\struct {G, *, \tau_G}$ be a topological group.
Let $g: \struct {S, \tau_S} \to \struct {G, \tau_G}$ be a continuous mapping.
Let $g^{-1}: S \to G$ be the mapping defined by:
- $\forall x \in S: \map {\paren {g^{-1} } } x = \paren {\map g x}^{-1}$
Then:
- $g^{-1}: \struct {S, \tau_S} \to \struct {G, \tau_G}$ is a continuous mapping.
Proof
By definition of a topological group:
- $\phi: \struct {G, \tau_G} \to \struct {G, \tau_G}$ such that $\forall x \in G: \map \phi x = x^{-1}$ is a continuous mapping
From Composite of Continuous Mappings is Continuous:
- the composition $\phi \circ g: \struct {S, \tau_S} \to \struct {G, \tau_G}$ is continuous.
Now:
\(\ds \forall x \in S: \, \) | \(\ds \map {\paren {g^{-1} } } x\) | \(=\) | \(\ds \paren {\map g x}^{-1}\) | Definition of $g^{-1}$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\map g x}\) | Definition of $\phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {\phi \circ g} } x\) | Definition of Composition of Mappings |
From Equality of Mappings:
- $g^{-1} = \phi \circ g$
The result follows.
$\blacksquare$