Epimorphism 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 an epimorphism.

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


Then $\struct {T, *}$ has the identity element $\map \phi {e_S}$.


Proof

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

Then:

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


Thus:

\(\displaystyle \map \phi x\) \(=\) \(\displaystyle \map \phi {x \circ e_S}\) $e_S$ is an identity for $\circ$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi x * \map \phi {e_S}\) Morphism property of $\circ$ under $\phi$

and:

\(\displaystyle \map \phi x\) \(=\) \(\displaystyle \map \phi {e_S \circ x}\) $e_S$ is an identity for $\circ$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {e_S} * \map \phi x\) Morphism property of $\circ$ under $\phi$

The result follows because every element $y \in T$ is of the form $\map \phi x$ with $x \in S$.

$\blacksquare$


Also see


Sources