Let $m\mid n$ and $\pi : \mathbb{Z}_n\rightarrow \mathbb{Z}_m$ be the canonical projection. Show that $\#\pi^{-1}(a) = n/m$. 
Let $m\mid n$ and $\pi : \mathbb{Z}_n\rightarrow \mathbb{Z}_m$ be the canonical projection given by $\overline r\mapsto \overline r$. Show that $\#\pi^{-1}(a) = n/m$, $a\in \mathbb{Z}_m$.

Here $\overline r$ means the equivalence class and $\#$ means cardinality. 
A quick example:
$\pi: \mathbb{Z}_{12}\rightarrow \mathbb{Z}_3$ we have $\pi^{-1}(0) = \{0,3,6,9\}$, $\pi^{-1}(1) = \{1,4,7,10\}$, $\pi^{-1}(2) = \{2,5,8,11\}$
which shows that in fact $\#\pi^{-1}(a) = 4=12/3 $ for every $a\in \mathbb{Z}_3$. 
I've managed to find an answer to this question, but it's not that much rigorous. Also i think that there may be a better proof for this fact. I hope you guys can help me improve my proof and if possible give a better solution to this. Here it goes:
In order to prove our result we prove the following:

Let $m\mid n$, $N=\{0,1,\dots,n-1\} $ and $R=\{0,1,\dots, m-1\}$. The quantity of elements $a\in N$ such that $ a\equiv r\pmod{m}$ is equal to $n/m$.

For each $r\in R$ we denote by $S_r= \{ mk+r:k \in \mathbb{N} \}$. We argue that $S_r\cap N$ goes for at maximum $k=\dfrac{n-m}{m}$. In fact, we note that $$n-m\leq m\bigg(\frac{n-m}{m}\bigg) + r\leq n-1$$
for every $r\in R$. Since $m\mid n$ it is clear that $n\leq m$ and therefore it is true that $n-m\in N$,  and also $n-1$ is clearly an element of $N$. This is the maximum value that $k$ can take, since if we consider: $k' = \dfrac{n-m}{m} +1$, this is actually equal to $\dfrac{n}{m}$ and hence $mk'+r = n+r \geq n \notin N $. 
Now, since $k$ varies from $0$ until $\dfrac{n-m}{m}$, we conclude that for each $r\in R$ there are $n/m$ elements.
Now we can proceed to the question itself. We simply note that for any $a\in \mathbb{Z}_m$, $\pi^{-1}(a) = \{\overline r+\alpha \overline m\}\cap\mathbb{Z}_n$ and since we are dealing with equivalence classes it is sufficient to deal only with the class representatives: $0,\dots, n-1$. By our lemma each $\pi^{-1}(a)$ contains exactly $n/m$ elements. 
 A: I think this could be solved also like this:
Firstly it can be directly computed that $$\#\pi^{-1}(0)=\frac{n}{m}.$$
Consider the left translation 
\begin{align}
L_a\colon \mathbb Z&\to\mathbb Z\\
x&\mapsto a+x
\end{align}
by $a\in\mathbb Z$, which also induces bijections $L_a\colon\mathbb Z_m\to\mathbb Z_m$ and $L_a\colon \mathbb Z_n\to\mathbb Z_n$ by $L_a(\overline x)=\overline a+\overline x$. Then for each $x+n\mathbb Z\in\mathbb Z_n$, it follows that
\begin{align}
&L_a\circ\pi(x+n\mathbb Z)\\
=&a+x+m\mathbb Z\\
=&(a+x)+m\mathbb Z\\
=&\pi(L_a(x+n\mathbb Z)).
\end{align}
I.e., the following diagram is commutative:
\begin{align}
\require{AMScd}
\begin{CD}
\mathbb Z_n @>{L_a}>> \mathbb Z_n\\
@V{\pi}VV @VV{\pi}V \\
\mathbb Z_m @>>{L_a}> \mathbb Z_m
\end{CD}
\end{align}
Now for each $a\in\mathbb Z$, for each  $x\in\pi^{-1}(a+n\mathbb Z)$, we have
\begin{align}
&\pi(x)=a+n\mathbb Z=L_a(0)\\
\Leftrightarrow &L_{-a}\circ\pi(x)=0\\
\Leftrightarrow &\pi\circ L_{-a}(x)=\pi(L_{-a}(x))=0\\
\Leftrightarrow &L_{-a}(x)\in\pi^{-1}(0)\\
\Leftrightarrow &x\in L_a(\pi^{-1}(0)),
\end{align}
which implies 
\begin{align}
\#\pi^{-1}(a+m\mathbb Z)=&\#L_a(\pi^{-1}(0))\\
=&\#\pi^{-1}(0)\\
=&\frac{n}{m}.
\end{align}

Or this fact can be proved very simply as follows.
For arbitrary $x,a\in\mathbb Z$, 
\begin{align}
&x+n\mathbb Z\in\pi^{-1}(a+m\mathbb Z)\\
\Leftrightarrow&\pi(x+n\mathbb Z)=a+m\mathbb Z\\
\Leftrightarrow& x+m\mathbb Z=a+m\mathbb Z\\
\Leftrightarrow& x-a\in m\mathbb Z\\
\Leftrightarrow& x-a+n\mathbb Z\in\pi^{-1}(0)\\
\Leftrightarrow& x+n\mathbb Z\in (a+n\mathbb Z)+\pi^{-1}(0)
\end{align}
Then $\pi^{-1}(a+m\mathbb Z)=(a+n\mathbb Z)+\pi^{-1}(0)$, while the coset $(a+n\mathbb Z)+\pi^{-1}(0)$ has the same cardinality as $\pi^{-1}(0)$ and thus $\#\pi^{-1}(a+m\mathbb Z)=n/m$.
A: If $f\colon G\to G'$ is a surjective group homomorphism, with $G$ and $G'$ finite groups, then
$$
f^{-1}(y)=x\ker f
$$
where $x\in G$ and $f(x)=y$ (prove it). In particular,
$$
|f^{-1}(y)|=|x\ker f|=\lvert\ker f\rvert
$$
(where $|X|=#X, if you prefer your notation).
In your case the only change is into additive notation. Thus you just need to compute $\lvert\ker\pi\rvert$. Can you do it?
