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 $\tau^*$ be the subspace topology on $R^*$.
By definition of a topological division ring:
- $\phi: \struct {R^*, \tau^*} \to \struct {R, \tau_{_R} }$ such that $\forall x \in R^*: \map \phi x = x^{-1}$ is a continuous mapping
Let $g^*: \struct {U, \tau_{_U} } \to \struct {R^*, \tau^*}$ 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:
\(\ds \forall x \in U: \, \) | \(\ds \map {\paren {g^{-1} } } x\) | \(=\) | \(\ds \map g x^{-1}\) | Definition of $g^{-1}$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\map g x}\) | Definition of $\phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\map {g^*} x}\) | Since $\map g x \ne 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {\phi \circ g^*} } x\) | Definition of Composition of Mappings |
From Equality of Mappings:
- $g^{-1} = \phi \circ g^*$
The result follows.
$\blacksquare$