Isomorphism Preserves Identity/Proof 1

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Theorem

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

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


Then $\circ$ has an identity $e_S$ if and only if $\map \phi {e_S}$ is the identity for $*$.


Proof

Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.

Then $\forall x \in S: x \circ e_S = x = e_S \circ x$.

The result follows directly from the morphism property of $\circ$ under $\phi$:

\(\ds \map \phi {x \circ e_S}\) \(=\) \(\ds \map \phi x\)
\(\ds \) \(=\) \(\ds \map \phi {e_S \circ x}\)
\(\ds \map \phi x * \map \phi {e_S}\) \(=\) \(\ds \map \phi x\)
\(\ds \) \(=\) \(\ds \map \phi {e_S} * \map \phi x\)


As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.

Thus the result for $\map \phi {e_S}$ can be applied to $\map {\phi^{-1} } {\map \phi {e_S} }$.

$\blacksquare$