Epimorphism Preserves Modules

Theorem
Let $\left({G, +_G, \circ}\right)_R$ be an $R$-module.

Let $\left({H, +_H, \circ}\right)_R$ be an $R$-algebraic structure.

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

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

It follows that the homomorphic image of an $R$-module is an $R$-module.

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

Proof
If $\left({G, +_G: \circ}\right)_R$ is an $R$-module, then:

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


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


 * $(3) \quad \left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$

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


 * $\forall x, y \in G: \phi \left({x +_G y}\right) = \phi \left({x}\right) +_H \phi \left({y}\right)$


 * $\forall x \in S: \forall \lambda \in R: \phi \left({\lambda \circ x}\right) = \lambda \circ \phi \left({x}\right)$


 * $\forall y \in H: \exists x \in G: y = \phi \left({x}\right)$

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, i.e. 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 criteria for $H$ being a module, in turn.


 * Module $(1)$:

Thus $(1)$ is shown to hold for $H$.


 * Module $(2)$:

Thus $(2)$ is shown to hold for $H$.


 * Module $(3)$:

Thus $(3)$ is shown to hold for $H$.

So all the criteria for $H$ to be an $R$-module are fulfilled.

Proof of Corollary

 * Let $G$ be a unitary $R$-module.

Then:


 * Module $(4) \quad \forall x \in G: 1_R \circ x = x$

So:

Thus $H$ is also a unitary module.