Showing $\phi: \mathbb Z \to \mathbb Z_{12}$ is a homomorphism and determining it's $Ker(\phi)$ and $Im(\phi)$ Question: Group Theory:  Let $\phi: \mathbb Z \to \mathbb Z_{12}$ be defined by $$\phi(x) = x\;(\text{mod}\; 12)$$ 
a). Show that $\phi$ is a homomorphism. 
My work: Let $\phi: \mathbb Z \to \mathbb Z_{12}$ be defined by $\phi(x)$ = $x$ $mod$ $12$
$\phi(x+y)$ $=$ $[x+y]_{12}$ $=$ $[x]_{12}$ $+$ $[y]_{12}$ $=$ $\phi(x)$ $+$ $\phi(y)$
$\therefore$ $\phi$ is a homomorphism.

b). Find $Ker(\phi)$ and $Im(\phi)$
My work: $Ker(\phi)$ $=$ {$x \in \mathbb Z$ $|   \phi(x) = [0]_{12}$}
                     $=$ {$x \in \mathbb Z$ $|$$[x]_{12} = [0]_{12}$}
                     $=$ {$x \in \mathbb Z$ $|$ $12$ divides $x$}
                     $=$ $12\mathbb Z$
$Im(\phi)$ $=$ {$\phi(x) \in \mathbb Z_{12} / x \in \mathbb Z$} $=$ $\mathbb Z_{12}$
Is my reasoning correct? I am unsure if my intuition is correct. Any help, comments, concerns, appreciated!

It is unclear what you asking since you've not explained 'where we are at'.
So the answer here will be general, forgetting all group theory and only assuming an understanding of equivalence relations.
Fact 1: The relation $x \equiv y \text{ (mod 12) }$ is an equivalence relation, which by definition means that $x - y$ is divisible by $12$.
The block containing $0$ is equal to the set $\{\dots,-36,-24,-12,0,12,24,36,\dots \}$; we denote this set by $12\mathbb Z$ and the quotient set by 
$\tag 1 \mathbb Z \text{/} 12\mathbb Z$
General theory tells us that we have a projection (or quotient) mapping
$\tag 2 \phi: \mathbb Z \to \mathbb Z \text{/} 12\mathbb Z$
The projection is by definition a surjective mapping.
Notice that $0$ gets mapped to $12\mathbb Z \in \mathbb Z \text{/} 12\mathbb Z$, and of course $\phi(x) =  12\mathbb Z$ just means that $x \in 12\mathbb Z $.
So this takes care of the questions on the image and the kernel without having to know any group theory.
