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:

$$\forall x, y, \in G, \forall \lambda, \mu \in R$$:
 * VS 1: $$\lambda \circ \left({x +_G y}\right) = \left({\lambda \circ x}\right) +_G \left({\lambda \circ y}\right)$$;
 * VS 2: $$\left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) +_G \left({\mu \circ x}\right)$$;
 * VS 3: $$\left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$$.

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.


 * VS 1:

$$ $$ $$ $$

Thus VS 1 is shown to hold for $$H$$.


 * VS 2:

$$ $$ $$ $$

Thus VS 2 is shown to hold for $$H$$.


 * VS 3:

$$ $$ $$ $$

Thus VS 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:


 * VS 4: $$\forall x \in G: 1_R \circ x = x$$.

So:

$$ $$

Thus $$H$$ is also a unitary module.