Preimage of Zero of Homomorphism is Submagma

From ProofWiki
Jump to navigation Jump to search

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:

\(\displaystyle x, y \in \phi^{-1} \sqbrk 0\) \(\leadstoandfrom\) \(\displaystyle \paren {\map \phi x = 0} \land \paren {\map \phi y = 0}\) Definition of Preimage of Element under Mapping
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle \map \phi x \circ \map \phi y = 0\) Definition of Zero Element
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle \map \phi {x * y} = 0\) Definition of Homomorphism (Abstract Algebra)
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle x * y \in \phi^{-1} \sqbrk 0\) Definition of Preimage of Element under Mapping

Hence the result.

$\blacksquare$