Inverse of Algebraic Structure Isomorphism is Isomorphism

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a mapping.


Then $\phi$ is an isomorphism if and only if $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is also an isomorphism.


General Result

Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ and $\struct {T, *_1, *_2, \ldots, *_n}$ be algebraic structures.

Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a mapping.


Then:

$\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ is an isomorphism

if and only if:

$\phi^{-1}: \struct {T, *_1, *_2, \ldots, *_n} \to \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ is also an isomorphism.


Proof

Let $\phi$ be an isomorphism.

Then by definition $\phi$ is a bijection.

Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.

That is:

$\exists \phi^{-1}: \struct {T, *} \to \struct {S, \circ}$


It follows that:

\(\ds \forall s \in S, t \in T: \, \) \(\ds \map \phi s\) \(=\) \(\ds t \iff \map {\phi^{-1} } t = s\) Inverse Element of Bijection
\(\ds \leadsto \ \ \) \(\ds \map \phi {s_1 \circ s_2}\) \(=\) \(\ds t_1 * t_2\) Definition of Morphism Property
\(\ds \leadsto \ \ \) \(\ds \map {\phi^{-1} } {t_1 * t_2}\) \(=\) \(\ds s_1 \circ s_2 = \map {\phi^{-1} } {t_1} \circ \map {\phi^{-1} } {t_2}\) Inverse Element of Bijection


So $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is a homomorphism.

$\phi^{-1}$ is also (from above) a bijection.

Thus, by definition, $\phi^{-1}$ is an isomorphism.


Let $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ be an isomorphism.

Applying the same result as above in reverse, we have that $\paren {\phi^{-1} }^{-1}: \struct {S, \circ} \to \struct {T, *}$ is also an isomorphism.

But by Inverse of Inverse of Bijection:

$\paren {\phi^{-1} }^{-1} = \phi$

and hence the result.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Inverse_of_Algebraic_Structure_Isomorphism_is_Isomorphism&oldid=686785"