We’ll first show the uniqueness. Let $h_{1},h_{2}\colon G/N\to H$ functions such that $h_{1}\circ\varphi=h_{2}\circ\varphi$. For an element $y$ in $G/N$ there exists an element $x$ in $G$ such that $\varphi(x)=y$, so we have

for all $y\in G/N$, thus $h_{1}=h_{2}$.

Now we define $h:G/N\to H,\;h(gN)=f(g)\;\forall\;g\in G$. We must check that the definition is independent of the given representative; so let $gN=kN$, or $k\in gN$. Since $N$ is a subset of $\ker(f)$, $g^{{-1}}k\in N$ implies $g^{{-1}}k\in\ker(f)$, hence $f(g)=f(k)$. Clearly $h\circ\varphi=f$.

Since $x\in\ker(f)$ if and only if $h(xN)=1_{H}$, we have

∎