Preimage of Submodule under Linear Transformation is Submodule

Theorem
Let $\struct {R, +_R, \times_R}$ be a ring.

Let $\struct {G, +_G, \circ_G}_R$ and $\struct {H, +_H, \circ_H}_R$ be $R$-modules.

Let $\phi: G \to H$ be a linear transformation.

Let $N$ be a submodule of $H$.

Then $\phi^{-1} \sqbrk N$ is a submodule of $G$.

Proof
Let $M = \phi^{-1} \sqbrk N$ be the preimage of $N$ under $\phi$.

$M$ is not a submodule of $G$.

This means that $M$ does not fulfil all the module axioms.

First suppose that:
 * $(1): \quad \struct {M, +_G}$ is not a subgroup of $G$.

Then $M$ is not a group.

Then by the One-Step Subgroup Test:
 * $\exists x, y \in M: x +_G \paren {-y} \notin M$

Then:

But by definition of $\phi$, both $\map \phi x \in N$ and $\map \phi {-y} \in N$.

Hence $\struct {N, +_G}$ is not closed.

By as applied to $N$, it follows that $N$ is not a group.

Hence $N$ is not a subgroup of $H$.

From this contradiction it follows that $(1)$ is false.

Thus $\struct {M, +_G}$ is a subgroup of $G$.

We have that $M$ is not a submodule of $G$.

We have shown that $\struct {M, +_G}$ is a subgroup of $G$.

Hence it must be the case that $M$ is not closed for scalar product:

But by definition of $\phi$, both $\map \phi x \in N$.

Hence $N$ is not closed for scalar product.

From this contradiction it follows that $(2)$ is false.

Thus $\struct {M, +_G}$ is closed for scalar product.

We have shown that $M$ is a subgroup of $G$ which is closed for scalar product.

That is, $M$ is a submodule of $G$.