If $f:G\to H$ is a homomorphism of groups, $N\lhd G$, $M\lhd H$, and $f(N)\lt M$, then $f$ induces a homomorphism $\bar{f}: G/N\to H/M$, given by $aN\mapsto f(a)M$. $\bar{f}$ is an isomorphism if and only if $\text{Im} f \vee M = H$ and $f^{-1}(M)\subset N$. In particular if $f$ is an epimorphism such that $f(N)= M$ and $\text{ker} f\subset N$, then $\bar{f}$ is an isomorphism.
Hungerford’s proof: Consider the composition $f:G\to H$ and $\pi :H\to H/M$ and verify that $N\subseteq f^{-1}(M)=\text{ker}\pi f$. By theorem 6 section 5 (applied to $\pi f$) the map $G/N\to H/M$ given by $aN\mapsto (\pi f)(a)=f(a)M$ is a homomorphism that is an isomorphism if and only if $\pi f$ is an epimorphism and $N = \text{ker} \pi f$. But the latter conditions hold if and only if $\text{Im} f\vee M=H$ and $f^{-1}(M)\subseteq N$. If $f$ is an epimorphism, then $H=\text{Im} f =\text{Im}f\vee M$. If $f(N)=M$ and $\text{ker}f\subseteq N$, then $f^{-1}(M)\subseteq N$, whence $\bar{f}$ is an isomorphism.
My attempt: We show $\bar{f}$ is a well defined function. If $aN=bN$, then $a^{-1}b\in N$. Since $f$ is homomorphism and $f(N)\subseteq M$, we have $f(a^{-1}b)=f(a)^{-1}f(b)\in M$. So $f(a)M=f(b)M$. Thus $\bar{f}(aN)= \bar{f}(bN)$ and $\bar{f}$ is well defined map. Let $aN,bN\in G/N$. Since $G/N$ and $H/M$ are group under multiplication $\cdot$ and $\circ$ respectively, and $f$ is homomorphism, we have $$\bar{f}(aN\cdot bN)=\bar{f}(abN)=f(ab)M=f(a)f(b)M=f(a)M\circ f(b)M=\bar{f}(aN)\bar{f}(bN).$$ Thus $\bar{f}$ is homomorphism.
We show $\bar{f}$ is an isomorphism if and only if $\text{Im} f \vee M = H$ and $f^{-1}(M)\subset N$. $(\Rightarrow)$ Suppose $\bar{f}$ is isomorphism. $\text{Im}f\vee M\subseteq H$ is trivial. Let $h\in H$. Then $hM\in H/M$. Since $\bar{f}$ is surjective, we have $\exists gN\in G/N$ such that $\bar{f}(gN)=f(g)M=hM$. So $f(g)^{-1}h\in M\subseteq \text{Im}f\vee M$. Since $f(g)^{-1}=f(g^{-1})\in \text{Im}f\vee M$, we have $h=f(g)f(g)^{-1}h\in \text{Im}\vee M$. Thus $H\subseteq \text{Im}f\vee M$. So $H=\text{Im}\vee M$. Let $x\in f^{-1}(M)$. Then $f(x)\in M$. So $\bar{f}(xN)=f(x)M=M$ and $xN\in \text{ker}\bar{f}$. Since $\bar{f}$ is injective, we have $\text{ker}\bar{f}=\{N\}$. So $xN=N$. Thus $x\in N$ and $f^{-1}(M)\subseteq N$. Hence $H=\text{Im}\vee M$ and $f^{-1}(M)\subseteq N$.
$(\Leftarrow)$ Suppose $\text{Im} f \vee M = H$ and $f^{-1}(M)\subset N$. Let $hM\in H/M$, where $h\in H$. Since $M\lhd H$ and $\text{Im}f\leq H$, we have $\text{Im}fM=\text{Im}f\vee M=M\text{Im}f$, by theorem 3 section 5. So $h=qm$, where $q\in \text{Im}f$ and $m\in M$. Thus $hM=qmH=qM=f(p)M=\bar{f}(pN)$. Hence $\exists pN\in G/N$ such that $\bar{f}(pN)=hM$. So $\bar{f}$ is surjective. Let $xN\in \text{ker}\bar{f}$. Then $\bar{f}(xN)=f(x)M=M$. So $f(x)^{-1}\in M$. Since $M\leq H$, we have $f(x)\in M$. That is $x\in f^{-1}(M)\subseteq N$. So $xN=N$. Thus $\text{ker}\bar{f}=\{N\}$ and $\bar{f}$ is injective. Hence $\bar{f}$ is isomorphism.
Suppose $f$ is an epimorphism, $f(N)= M$ and $\text{ker} f\subset N$. Let $hM\in H/M$, where $h\in H$. Since $f$ is surjective, $\exists g\in G$ such that $f(g)=h$. So $\exists gN\in G/N$ such that $\bar{f}(gN)=f(g)M=hM$. Thus $\bar{f}$ is surjective. Let $xN\in \text{ker}\bar{f}$. Then $\bar{f}(xN)=f(x)M=M$. So $f(x)\in M=f(N)$. Then $\exists n\in N$ such that $f(x)=f(n)$. Since $f$ is homomorphism, we have $f(xn^{-1})=f(x)f(n)^{-1}=e_H$. So $xn^{-1}\in \text{ker}f\subseteq N$. Which implies $x=xn^{-1}n\in N$ and $xN=N$. Thus $\text{ker}\bar{f}=\{N\}$. So $\bar{f}$ is injective. Hence $\bar{f}$ is isomorphism. Is my proof correct?
In last paragraph, I’m unsure of my proof of injectivity of $\bar{f}$. I think in following proof we don’t use $\text{ker}f\subseteq N$ hypothesis: Let $xN\in \text{ker}\bar{f}$. Then $\bar{f}(xN)=f(x)M=M$. So $f(x)\in M=f(N)$. Which implies $x\in N$. Thus $xN=N$ and $\bar{f}$ is injective.
Is proof of $\pi f$ is an epimorphism$\iff$$\text{Im}f \vee M=H$ trivial or similar to mine proof (which is kind of long)? If it is similar to mine, then author didn’t want to add details of proof.
Reading your comments, I have an idea for that $πf$ is an epimorphism.
My attempts: $${\rm Im}\pi{f}=H/M \Longleftrightarrow\pi{\rm Im}{f}=H/M\Longleftrightarrow{{\rm Im}fM}=H$$ Since ${\rm Im}{f}$ is a subgroup and $M$ is a normal group, so we get that $${\rm Im}f\vee{M}=H$$