Epimorphism Preserves Semigroups

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism.

Let $\left({S, \circ}\right)$ be a semigroup.


Then $\left({T, *}\right)$ is also a semigroup.


Proof

As $\left({S, \circ}\right)$ is a semigroup, then by definition it is closed.

As $\phi$ is an epimorphism, it is by definition surjective.

That is:

$T = \phi \left[{S}\right]$

where $\phi \left[{S}\right]$ denotes the image of $S$ under $\phi$.

From Morphism Property Preserves Closure it follows that $\left({T, *}\right)$ is closed.


As $\left({S, \circ}\right)$ is a semigroup, then by definition $\circ$ is associative.

From Epimorphism Preserves Associativity, $*$ is therefore also associative.


So:

$\left({T, *}\right)$ is closed

and:

$*$ is associative.

Therefore, by definition, $\left({T, *}\right)$ is a semigroup.

$\blacksquare$


Warning

Note that this result is applied to epimorphisms.

For a general homomorphism which is not surjective, nothing definite can be said about the behaviour of the elements of its codomain which are not part of its image.


Also see


Sources