Proof that $\phi (m) = m \pmod{n}$ is a homomorphism from $ \mathbb Z$ to $ \mathbb Z_n$

Question

I found this example in Contemporary Abstract Algebra by J. Gallian.

It was given that

$\phi (m) = m\pmod{n}$ is a homomorphism from $ \mathbb Z$ to $\mathbb Z_n$

I wanted to prove it and was able to reach to the following steps

$\phi(a+b) = (a+b)\pmod{n}$
= $(a\pmod{n}+ b\pmod{n})\pmod{n} $

Since $\phi(a) = a\pmod{n}$

Therefore, $\phi(a+b) = (\phi(a)+\phi(b))\pmod{n} $

I'm not sure what should be next steps so that we obtain.

$$\phi(a+b) = \phi(a) + \phi(b)$$

Edit Update: I was able to figure out the Answer it's mentioned as an answer for future reference.

Answer 1

So the thing was, I misunderstood the binary operation in $ \Bbb Z_{n} $ to be precise, $$ a*b = (a+b)(mod\,n)\;\; in \;\; \Bbb Z_{n} $$ Therefore, $$ \phi(a) * \phi(b) = (\phi(a) + \phi(b))(mod\,n)\,,\;\;\; where\;\;\phi(a) , \phi(b) \in \Bbb Z_{n} $$

And this proves that the mentioned map is a homomorphism.

Answer 2

You have$$a+b-(a\operatorname{mod}n+b\operatorname{mod}n)=a-a\operatorname{mod}n+b-b\operatorname{mod}n\tag1$$and each of the numbers $a-a\operatorname{mod}n$ and $b-b\operatorname{mod}n$ is a multiple of $n$. So, the LHS of $(1)$ is a multiple of $n$. Therefore, $(a+b)\operatorname{mod}n=\bigl(a\operatorname{mod}n+b\operatorname{mod}n\bigr)\operatorname{mod}n$.

Answer 3

$\phi(a+b) = (\phi(a)+\phi(b))\pmod n$

gives $\phi(a+b) = \phi(a)\pmod n+\phi(b)\pmod n = \phi(\phi(a))+\phi(\phi(b))$.

So you need to prove $\phi$ is idempotent, i.e. $\phi(\phi(m))=\phi(m)$.