Let $\phi:G\rightarrow K$ be a group homomorphism. Suppose $N$ is a normal subgroup of $G$. Show that if $\psi:G/N\rightarrow K$ defined by $\psi(gN)=\phi(g)$ for all $g\in G$ is well-defined, then $N\subset \ker\phi$.
Attempt: If we take $g_1N=g_2N$, then $g_1^{-1}g_2\in N$ and $\phi(g_1)=\phi(g_2)$. This means $\phi(g_1^{-1}g_2) = \phi(g_1)^{-1}\phi(g_2)=e$. So $g_1^{-1}g_2\in\ker\phi$.
But using this approach, I cannot guarantee $g_1^{-1}g_2$ covers all elements in $N$.