Homomorphic Image of Group Element is Coset

Theorem
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.

Let $\map \ker \phi$ be the kernel of $\phi$.

Let $h \in H$.

Then $\Preimg h$ is either the empty set or a coset of $\map \ker \phi$.

Proof
There are two possibilities for any $h \in H$.
 * $(1): \quad \Preimg h = \O$
 * $(2): \quad \Preimg h \ne \O$

If $(1)$, then the first disjunct of the result is satisfied.

Now suppose $(2)$ holds.

Let $e_G$ and $e_H$ be the identity elements of $G$ and $H$ respectively.

Let $K = \map \ker \phi$.

Let $x, y \in G$ such that $\map \phi x = \map \phi y$.

Then:

Thus:
 * $\set {x: \map \phi x = \map \phi y}$ is a subset of $y K$.

From Kernel is Normal Subgroup of Domain we have that $y K = K y$.

Now suppose $x \in K y$.

Then, by definition, $x = k y$ for some $k \in K$.

Thus:

A similar process gives that $x \in y K \implies \map \phi x = \map \phi y$.

Hence the result.