Preimage of Zero of Homomorphism is Submagma
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Theorem
Let $\struct {S, *}$ be a magma.
Let $\struct {T, \circ}$ be a magma with a zero element $0$.
Let $\phi: S \to T$ be a magma homomorphism.
Then $\struct {\phi^{-1} \sqbrk 0, *}$ is a submagma of $\struct {S, *}$.
Proof
Let $x, y \in \phi^{-1} \sqbrk 0$.
It is to be shown that:
- $x * y \in \phi^{-1} \sqbrk 0$
Thus:
\(\ds x, y \in \phi^{-1} \sqbrk 0\) | \(\leadstoandfrom\) | \(\ds \paren {\map \phi x = 0} \land \paren {\map \phi y = 0}\) | Definition of Preimage of Element under Mapping | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \map \phi x \circ \map \phi y = 0\) | Definition of Zero Element | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \map \phi {x * y} = 0\) | Definition of Homomorphism (Abstract Algebra) | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds x * y \in \phi^{-1} \sqbrk 0\) | Definition of Preimage of Element under Mapping |
Hence the result.
$\blacksquare$