Preimage of Normal Subgroup under Epimorphism is Normal Subgroup

Theorem
Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.

Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group epimorphism.

Let $H$ be a normal subgroup of $\struct {G_2, *}$.

Then:
 * $\phi^{-1} \sqbrk H$ is a normal subgroup of $\struct {G_1, \circ}$

where $\phi^{-1} \sqbrk H$ denotes the preimage of $H$ under $\phi$.

Proof
Let $H$ be a normal subgroup of $\struct {G_2, *}$.

First note that from Preimage of Subgroup under Epimorphism is Subgroup:
 * $\phi^{-1} \sqbrk H$ is a subgroup of $\struct {G_1, \circ}$.

It remains to be shown that $\phi^{-1} \sqbrk H$ is normal.

Let $K = \phi^{-1} \sqbrk H$.

That is:
 * $H = \map \phi K$

Let $g \in G$ be arbitrary.

To prove that $K$ is normal, we wish to show that:
 * $\forall g \in G: g \circ K = K \circ g$

We have:
 * $g \circ K = \set {y \in G: \exists k \in K: g \circ k = y}$

So, let $k \in K$ be arbitrary.

Let $\map \phi k = h$.