# Preimage of Zero of Homomorphism is Submagma

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.

$\blacksquare$