Combination Theorem for Continuous Mappings/Topological Group/Inverse Rule

From ProofWiki
Jump to navigation Jump to search

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

\(\, \displaystyle \forall x \in S : \, \) \(\displaystyle \map {\paren {g^{-1} } } x\) \(=\) \(\displaystyle \paren {\map g x}^{-1}\) Definition of $g^{-1}$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {\map g x}\) Definition of $\phi$
\(\displaystyle \) \(=\) \(\displaystyle \map {\paren {\phi \circ g} } x\) Definition of Composition of Mappings

From Equality of Mappings:

$g^{-1} = \phi \circ g$

The result follows.

$\blacksquare$