Show $f:\mathbb{Z}_6 \rightarrow \mathbb{Z}_3$ is a homomorphism I am currently studying for an abstract algebra final exam. I am trying to disprove the statement "Consider the homomorphism $f: R \rightarrow S$ where R and S are rings. Prove/Disprove: If $a \in R$ is not a unit in $R$, then $s = f(a)$ is NOT a unit in S.
I know this is false because taking  $f:\mathbb{Z}_6 \rightarrow \mathbb{Z}_3$,
then f looks like $f([a]_6) = [a]_3$. Then if we let $a = 4$, a is not a unit in $\mathbb{Z}_6$ because it's a zero divisor. However, $a$ IS a unit in $\mathbb{Z}_3$, because $[4]_3 = [1]$, which is a unit.
My trouble is coming in where proving $f:\mathbb{Z}_6 \rightarrow \mathbb{Z}_3$ is indeed a homomorphism.
Here is my attempt:
Let $a, b \in R$. So $a = 6n + r_1$ and $b = 6m + r_2$ for $n, m, r_1, r_2 \in \mathbb{Z}$. So
$f([a]_6 + [b]_6) = f(a) + f(b)$
$f([6n + r_1 + 6m + r_2]_6) = [6n + r_1]_3 + [6m + r_2]_3$ 
After this line, I'm not exactly sure how to convert the elements in $\mathbb{Z}_6$ to look like elements in $\mathbb{Z}_3$. 
If anyone could help me out that would be greatly appreciated.
 A: Remember to also check that $f$ is well-defined upto choice of representative. Here is how I would write the proof:
Fix $a + 6\mathbb{Z}, b + 6\mathbb{Z} \in \mathbb{Z}/6\mathbb{Z}$. Then, 
$$
\begin{align}
f((a + 6\mathbb{Z}) + (b + 6\mathbb{Z})) &= f((a + b) + 6\mathbb{Z}) \\
&= (a + b) + 3\mathbb{Z} \\
&= (a + 3\mathbb{Z}) + (b + 3\mathbb{Z}) \\
&= f(a + 6\mathbb{Z}) + f(b + 6\mathbb{Z})
\end{align}
$$
You can similarly show that $1$ goes to $1$ and that multiplication is preserved (although the latter follows from the fact that addition is preserved).
A: Your function is given by $f(a)=a\bmod 3$. And you are asking whether
$$f(a+_6b)=f(a)+_3f(b).$$
The two operations are given by $a+_6b=(a+b)\bmod 6$ (on $\mathbb Z_6=\{0,1,2,3,4,5\}$ and $a+_3b=(a+b)\bmod3$ (on $\mathbb Z_3=\{0,1,2\}$. So this is equivalent to the question whether
$$((a+b)\bmod 6)\bmod 3 = (a\bmod 3+b\bmod 3)\bmod 3.$$
You should be able to show that
\begin{align*}
((a+b)\bmod 6)\bmod 3 &= (a+b)\bmod 3\\
(a\bmod 3+b\bmod 3)\bmod 3 &= (a+b)\bmod 3
\end{align*}
Just notice that the two numbers give the same remainder modulo $3$ if their difference is a multiple of $3$, i.e., 
$$x\bmod 3=y\bmod 3 \qquad\Leftrightarrow\qquad 3\mid x-y.$$
Which is the same as saying that $x$ and $y$ are congruent modulo $3$, if you are familiar with this notion.
So all you can to do is check whether $(a+b)-((a+b)\mod 6)$ and $(a+b)-(a\bmod 3+b\bmod 3)$ are multiples of $3$.
You can use similar arguments to show that $f$ preserves also multiplication, i.e., that $f(a\cdot_6 b)=f(a)\cdot_6 f(b)$. (And, as I mentioned, the whole problem becomes much simpler if you are familiar with basic properties of congruences and modular arithmetic.)
