Edit: I added my own attempt at an answer after working some things out. Could anyone check if the answer I gave is correct?
I was given the following question:
Let $G$ be a group, $N_1,N_2\triangleleft G$. Define the function $f:G\to (G/N_1)\times (G/N_2)$ by $f(a)=(aN_1, aN_2).$ Prove that $f$ is a group-homomorphism with kernel $N_1\cap N_2$.
I proceeded to try some things, but I find it hard to prove things when the operations of the groups are not given. I thought that the following would work, but the lack of concrete operations leaves me on defense:
$f(ab)=(abN_1, abN_2) = (aN_1bN_1, aN_2bN_2) = (aN_1, aN_2)(bN_1, bN_2) = f(a)f(b)$.
This would mean that some sort of product $\circ$ would be defined on $(G/N_1)\times (G/N_2)$ by $(a, b)\circ(x,y) = (ax,by)$, but again I can't really find this anywhere concretely. Is this all correct?
For the kernel part: am I looking for $N(f)=\{a\in G: f(a)=(0N_1,0N_2) = (\bar{0}, \bar{0})\}$?
$f$ is a group homomorphism:
$f(ab)=(abN_1, abN_2) = (aN_1bN_1, aN_2bN_2) = (aN_1, aN_2)(bN_1, bN_2) = f(a)f(b)$.
$N(f)=N_1\cap N_2$
(1) Suppose $a\in N_1\cap N_2$, then $aN_1=N_1=\bar{e}$ and $aN_2=N_2=\bar{e}$, so that we have $f(a)=(\bar{e}, \bar{e}).$ So $a\in N(f)$ which means $N_1\cap N_2\subset N(f)$.
(2) Suppose $a\in N(f)$, then $f(a)=(\bar{e}, \bar{e})= (N_1, N_2) = (aN_1, aN_2)$.
Suppose that $a\in N_1\backslash N_2$, then $aN_1=\bar{e}$ but $aN_2\neq \bar{e}$ which contradicts the fact that $f(a)=(\bar{e}, \bar{e})= (N_1, N_2) = (aN_1, aN_2)$. The same contradiction follows from $a\in N_2\backslash N_1$.
So, $a$ must be in both $N_1$ and $N_2$, which gives us $N(f)\subset N_1\cap N_2$.
From (1) and (2) we can conclude that $N(f)=N_1\cap N_2$.