$\mathbb{Z}_m$ is homomorphic image of $\mathbb{Z}_n$ 
Doesn't this always work as long as $n\geq m$? Can't we get rid of the condition that $n$ is a multiple of $m$?

If $n$ is a multiple of $m$, show that $\mathbb{Z}_m$ is homomorphic image of $\mathbb{Z}_n$ (ring homomorphism).
We define $f:\mathbb{Z}_n\rightarrow\mathbb{Z}_m$ so that $f(a)=a\pmod m$. Then $f(a+b)=f(a)+f(b)$ and $f(ab)=f(a)f(b)$. 
 A: Try $f : \Bbb Z_3 \to \Bbb Z_2$.
Observe that in $\mathbb Z_3$, $2 + 2 = 1$. So $f(2 + 2) = f(1)$.
What is $f(2) + f(2)$?
(I'd argue that the real problem here is in your definition of $f$. You write $f(a) = a \pmod m$, but what is $a$? It is an argument of $f$, so it should be an element of $\mathbb Z_m$, but then what does it mean to take such a thing mod $n$? Normally in $x \pmod n$, $x$ is an integer, and if you're going to extend this to $a \in \mathbb Z_m$ by picking some integer congruent to $a$, you had better show that it doesn't matter which one you choose).
A: Let $f:\Bbb Z_3\to\Bbb Z_2$. 
$0=f(0)=f(1+2)=f(1)+f(2)=1+0=1$
This example show that the $f$ you defined is not homomorphism from $\mathbb Z_3\to\mathbb Z_2$
A: Assume that $\def\Z{\Bbb {Z}}f\colon\Z_n\to\Z_m$ is a surjective group homomorphism. Then $H=\ker f$ is a subgroup of $\Z_n$ having index $m$. So, by Lagrange's theorem,
$$
n=|\Z_n|=[\Z_n:H]\cdot|H|=m|H|
$$
which means that $m$ divides $n$.
If in your definition of ring homomorphism it's required that the unit goes to the unit, the fact that (the class of) $1$ generates $\Z_k$ as a group with respect to addition (for all $k$) implies that a ring homomorphism $\Z_n\to\Z_m$ is surjective (and also uniquely defined).
