If $f:G\to H$ is a homomorphism of groups and $N$ is a normal subgroup of $G$ contained in the kernel of $f$, then there is a unique homomorphism $\bar{f}: G/N\to H$ sucht that $\bar{f}(aN)= f(a)$ for all $a\in G$. $\text{Im} f= \text{Im} \bar{f}$ and $\text{ker}\bar{f}= (\text{ker}f)/N$. $\bar{f}$ is an isomorphism if and only if $f$ is an epimorphism and $N=\text{ker} f$.
Hungerford’s proof: If $aN=bN$, then $ab^{-1}\in N$. Since $f$ is homomorphism and $N\subseteq \text{ker}f$, we have $f(ab^{-1})=f(a)f(b)^{-1}=e_H$. So $f(a)=f(b)$. Thus $\bar{f}(aN)=f(a)=f(b)=\bar{f}(bN)$ and $\bar{f}$ is a well defined function. Since $\bar{f}(aNbN) = \bar{f}(abN) = f(ab)=f(a)f(b)= \bar{f}(aN) \bar{f}(bN)$, $\bar{f}$ is a homomorphism. Clearly $\text{Im}\bar{f}= \text{Im}f$ and $$aN\in \text{ker}\bar{f}\Leftrightarrow f(a)=e \Leftrightarrow a\in \text{ker}f,$$ whence $\text{ker}\bar{f}=\{aN\mid a\in \text{ker} f\} = (\text{ker} f)/ N$. $\bar{f}$ is unique since it is completely determined by $f$. Finally it is clear that $\bar{f}$ is an epimorphism if and only if $f$ is. By Theorem 2.3 $\bar{f}$ is a monomorphism if and only if $\text{ker}\bar{f}= (\text{ker}f)/N$ is the trivial subgroup of $G/N$, which occurs if and only if $\text{ker} f = N$.
Question: I don’t understand uniqueness part in above theorem and its proof “$\bar{f}$ is unique since it is completely determined by $f$“. Does it say, $\exists !\ \bar{f}:G/N\to H$ such that $\bar{f}$ is homomorphism, $\text{Im}f=\text{Im}\bar{f}$ and $\text{ker}\bar{f}=(\text{ker}f)/N$?
The first part proves that there exists a homomorphism $\bar f$ with $\bar f(aN)=f(a)$ for all $a\in G$. The uniqueness part says that if $g$ is another homomorphism with $g(aN)=f(a)$ for all $a\in G$, then $g=\bar f$.
The phrase ‘$\bar f$ is completely determined by $f$’ is saying that the condition $\bar f(aN)=f(a)$ already determines $\bar f$ as a map of sets, so there is no other choice for $g$.