Homomorphism with Cancellable Codomain Preserves Identity

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Theorem

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

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

Let $\struct{S, \circ}$ have an identity $e_S$.

Let $\struct{T, *}$ have an identity $e_T$.

Let every element of $\struct{T, *}$ be cancellable.


Then $\map \phi {e_S}$ is the identity $e_T$.


Proof

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

Let $\struct{T, *}$ be an algebraic structure in which $*$ has an identity $e_T$.

Let $\struct{T, *}$ be such that every element is cancellable.

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


Every element of $\struct{T, *}$ is cancellable.

Suppose there is an idempotent element in $\struct{T, *}$

So from Identity is only Idempotent Cancellable Element, it must be the identity $e_T$.

Thus:

\(\displaystyle \map \phi {e_S} * \map \phi {e_S}\) \(=\) \(\displaystyle \map \phi {e_S \circ e_S}\) by the morphism property of $\circ$ under $\phi$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {e_S}\) as $e_S$ is the identity of $\left({S, \circ}\right)$


So $\map \phi {e_S}$ is idempotent in $\struct{T, *}$ and the result follows.

$\blacksquare$


Also see


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