Rules on surjective and Injective

Question

So I am trying to prove $f([a]+[b]) = f([a]) + f([b])$ for $\mathbb Z$ mod $12$ and the same for multiplication... Can you show that this is true based on a function being surjection and injective and any combination of the two without showing every possibility of every element in $\mathbb Z$ mod $12$? Is there any definitions stating this?

Answer 1

So, if I understand right, you are given $f(x)=3x$, defined on $\Bbb Z_{12}$.

This maps $0\mapsto 0,\ 1\mapsto 3,\ 2\mapsto 6,\ 3\mapsto 9,\ 4\mapsto 0,\ \dots$.

We can say that $f:\Bbb Z_{12}\to\Bbb Z_{12}$, in this case it is neither injective nor surjective. (Why?)

The range of $f$ is $im(f)=\{0,3,6,9\}\subseteq\Bbb Z_{12}$ which is isomorphic to $\Bbb Z_4=\{0,1,2,3\}$.

So, we can also write $f:\Bbb Z_{12}\to\, im(f)\ \cong \Bbb Z_4$, and in this sense $f$ is going to be surjective.

And, finally, the fact that $f$ preserves addition follows simply by linearity: $$f(a+b)\ =\ 3(a+b)\ =\ 3a\ +\ 3b\ =\ f(a)\,+\,f(b)\,.$$