I defined a function from two rings $f$: $Z_{mn} \mapsto Z_{m}$ $\times$ $Z_{n}$ as $f(x\pmod{mn}) = (x\pmod{m}, x\pmod{n})$ and I want to show that it is a homomorphism.
Observe that for any $x,y \in Z_{mn}, f(x)= (x\pmod{m}, x\pmod{n})$. Similarly, $f(y)= (y\pmod{m}, y\pmod{n})$.
Now, $f(x)+f(y)=(x\pmod{m},x\pmod{n})+(y\pmod{m},y\pmod{n})$ (1)
= $(x\pmod{m}+y\pmod{m}),(x\pmod{n}+y\pmod{n})$ (2)
= $(x+y\pmod{m}),(x+y\pmod{n})$ (3)
= $f(x+y)$. (4)
Similarly for multiplication.
Observe that for any $x,y \in Z_{mn}, f(x)= (x\pmod{m}, x\pmod{n})$. Similarly, $f(y)= (y\pmod{m}, y\pmod{n})$.
Now, $f(x)f(y)=(x\pmod{m},x\pmod{n})*(y\pmod{m},y\pmod{n})$ (1)
= $(x\pmod{m}*y\pmod{m}),(x\pmod{n}+y\pmod{n})$ (2)
= $(xy\pmod{m}),(xy\pmod{n})$ (3)
= $f(xy)$. (4)
However, my book says that $x\pmod{n} + y\pmod{n} = (x+y\pmod{n})\pmod{n}$ and $x\pmod{n} * y\pmod{n} = (xy\pmod{n})\pmod{n}$. But this is not what I need to justify step (2) to (3) so what am I doing wrong? Thank you.