Combination Theorem for Continuous Mappings/Topological Division Ring/Inverse Rule

Theorem
Let $\struct{S, \tau_S}$ be a topological space.

Let $\struct{R, +, *, \tau_R}$ be a topological division ring.

Let $g : \struct{S, \tau_S} \to \struct{R, \tau_R}$ be a continuous mapping.

Let $U = S \setminus \set{x : \map g x = 0}$

Let $g^{-1} : U \to R$ be the mapping defined by:
 * $\forall x \in U : \map {g^{-1}} x = \map g x^{-1}$

Let $\tau_U$ be the subspace topology on $U$.

Then
 * $g^{-1} : \struct{U, \tau_U} \to \struct{R, \tau_R}$ is continuous.

Proof
Let $R^* = R \setminus \set{0}$.

Let $\tau_{R^*}$ be the subspace topology on $R^*$.

By definition of a topological division ring:
 * $\phi: \struct{R^*, \tau_{R^*}} \to \struct{R, \tau_R}$ such that $\forall x \in R^*: \map \phi x = x^{-1}$ is a continuous mapping

Let $g^* : \struct{U, \tau_U} \to \struct{R^*, \tau_{R^*}}$ be the restriction of $g$ to $U \subseteq R$.

From Restriction of Continuous Mapping is Continuous, $g^*$ is a continuous mapping.

From Composite of Continuous Mappings is Continuous, the composition $\phi \circ g^* : \struct{U, \tau_U} \to \struct{R, \tau_R}$ is continuous.

Now

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

The result follows.