# 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:

- $\forall x \in S: \map {\paren{g^{-1}}} x = \map g x^{-1}$

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

\(\, \displaystyle \forall x \in S : \, \) | \(\displaystyle \map {\paren {g^{-1} } } x\) | \(=\) | \(\displaystyle \map g x^{-1}\) | 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.

$\blacksquare$