Epimorphism Preserves Modules

Theorem
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.

Let $\struct {H, +_H, \circ}_R$ be an $R$-algebraic structure.

Let $\phi: G \to H$ be an epimorphism.

Then $H$ is an $R$-module.

Corollary
If $G$ is a unitary $R$-module, then so is $H$.

Proof
If $\struct {G, +_G, \circ}_R$ is an $R$-module, then:

$\forall x, y, \in G, \forall \lambda, \mu \in R$:
 * $(1): \quad \lambda \circ \paren {x +_G y} = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$


 * $(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$


 * $(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$

If $\phi: G \to H$ is an epimorphism, then:


 * $\forall x, y \in G: \map \phi {x +_G y} = \map \phi x +_H \map \phi y$


 * $\forall x \in S: \forall \lambda \in R: \map \phi {\lambda \circ x} = \lambda \circ \map \phi x$


 * $\forall y \in H: \exists x \in G: y = \map \phi x$

As $\phi$ is an epimorphism, we can accurately specify the behaviour of all elements of $H$, as they are the images of elements of $G$.

If $\phi$ were not an epimorphism, that is not surjective, we would have no way of knowing the behaviour of elements of $H$ outside of the image of $G$.

Hence the specification that $\phi$ needs to be an epimorphism.

Now we check the module axioms in turn.

Thus is shown to hold for $H$.

Thus is shown to hold for $H$.

Thus is shown to hold for $H$.

So all the module axioms for $H$ are satisfied.

Also see

 * Homomorphic Image of R-Module is R-Module