Homomorphic Image of Group Element is Coset

Theorem
Let $\phi: \left({G, \circ}\right) \to \left({H, *}\right)$ be a group homomorphism.

Let $\ker \left({\phi}\right)$ be the kernel of $\phi$.

Let $h \in H$.

Then $\operatorname{Im}^{-1} \left({h}\right)$ is either the empty set or a coset of $\ker \left({\phi}\right)$.

Proof
There are two possibilities for any $\phi(g) \in H$.
 * $(1): \quad \operatorname{Im}^{-1} \left[ \; \{ \; \phi(g) \; \} \; \right] = \varnothing$
 * $(2): \quad \operatorname{Im}^{-1} \left[ \; \{ \; \phi(g) \; \} \; \right] \ne \varnothing$

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.

1. We prove $\left\{{x \in G : \phi \left({x}\right) = \phi \left({y}\right)}\right\} \subseteq y \ker {\phi} $
Let $x, y \in G$ such that $\phi \left({x}\right) = \phi \left({y}\right)$.

Then:

Thus:
 * $\left\{{x \in G : \phi \left({x}\right) = \phi \left({y}\right)}\right\}$ is a subset of $y \ker {\phi}$.

From Kernel is Normal Subgroup of Domain we have that $y \ker {\phi} = (\ker {\phi}) y$.

2. We prove $(\ker {\phi})y \subseteq \left\{{x \in G : \phi \left({x}\right) = \phi \left({y}\right)}\right\} $
Now suppose $x \in (\ker {\phi}) y$.

Then, by definition, $x = k y$ for some $k \in \ker {\phi}$.

Thus:

3. We prove $y \ker {\phi} \subseteq \left\{{x \in G : \phi \left({x}\right) = \phi \left({y}\right)}\right\} $
A similar process to 2. gives that $x \in y \ker {\phi} \implies \phi \left({x}\right) = \phi \left({y}\right)$.

Hence the result.