Combination Theorem for Continuous Mappings/Topological Group/Inverse Rule

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

From Equality of Mappings, $g^{-1} = \phi \circ g$.

The result follows.