Prove that $\Bbb{Z}/12\Bbb{Z}\to\Bbb{Z}/4\Bbb{Z}$ is a homomorphism Prove that $f:\Bbb{Z}/12\Bbb{Z}\to\Bbb{Z}/4\Bbb{Z}:[x]_{12}\mapsto[x]_4$ is a homomorphism.
I think you can prove this very simply using the following reasoning:
\begin{align*}
[x]_{12},[y]_{12}\in\Bbb{Z}/12\Bbb{Z}\implies f([x]_{12}+[y]_{12}) &= f([x+y]_{12})\\ 
 &=[x+y]_4 \\ 
 &=[x]_4+[y]_4 \\ 
 &=f([x]_{12})+f([x]_{12}) 
\end{align*}
But I was thinking that I may need to prove that $f$ is a well-defined function, otherwise this (homework) exercise seems a little bit too easy. 
Should I prove this ? And any hints how to do this ?
 A: It might appear a bit finnicky, but until you have shown that the function is well defined, you cannot use the function symbol, because it might be ambiguous.
So you have to prove that if $[x]_{12} = [y]_{12}$ for two integers $x, y$, then $[x]_{4} = [y]_{4}$. The first equality means $12 \mid x -y$, the second one $4 \mid x -y$. Since $4 \mid 12$, and divisibility is transitive, you get the first equality implies the second one.
A: The idea is that if $x$ and $y$ are such that $[x]_{12}=[y]_{12}$, then $f([x]_{12})=f([y]_{12})$ because $4$ divides $12$.
Indeed, there exist $k \in \mathbb{Z}$ such that $y=x+12k$, so $f([y]_{12})=[y]_4=[x+12k]_4=[x]_4=f([x]_{12})$.
A: Suppose $x\equiv y \mod 12$. Then $x-y=12k=4(3k)=4m$ so $x\equiv y\mod 4$. It is well defined.
In general, $a\equiv b \mod mn\implies a\equiv b\mod m,\mod n$, but not conversely.
A: The key point is that $4$ divides $12$ or, equivalently $12\mathbb{Z}\subseteq 4\mathbb{Z}$.
Also, note that this is a ring homomorphism.
I hear already some people tell me that you probably have not seen this yet, but I will still propose you an alternative that could be interesting, now or later.
This has to do with the factorization of homomorphisms through a quotient. The condition is that the kernel contains the ideal, subgroup, subspace, etc... by which you want to mod out the domain of the initial homomorphism. This a general and useful principle. 
First note that the canonical surjection
$$
x\longmapsto [x]_{4}
$$
is a (well-defined, of course) homomorphism from $\mathbb{Z}$ onto $\mathbb{Z}/4\mathbb{Z}$.
Now observe that the kernel of the latter is $4\mathbb{Z}$ and contains $12\mathbb{Z}$.
It follows that the above homomorphism factors through $\mathbb{Z}/12\mathbb{Z}$ into a homomorphism
$$
[x]_{12}\longmapsto [x]_{4}.
$$
See theorem 2.1 (for group homomorphisms, but the same is true for ring homorphisms through quotients modulo ideals like here) here: http://www.math.rochester.edu/people/faculty/jonpak/N4.pdf
