Proving that two quotient modules are isomorphic Given a ring $R$ and $R$-modules $A,B,C,D$ such that $$\sigma:A \rightarrow B, \tau: C \rightarrow D, \rho: A \rightarrow C, \kappa: B \rightarrow D, \ \mathrm{and} \ \kappa \circ \sigma = \tau \circ \rho,$$ where $\sigma, \tau, \rho, \kappa$ are homomorphisms and $\rho, \kappa$ are isomorphisms, 
I want to show that $\dfrac{B}{\mathrm{Im}{\sigma}} \cong \dfrac{D}{\mathrm{Im}{\tau}}.$
I have proven the following result, and am applying it to solve the given problem.

Result:
Let $R$ be a ring and let $B,D$ be two isomorphic $R$-modules. Let $\mathrm{Im}{\sigma} \subseteq B$ be a submodule of $B$ and $\mathrm{Im}{\tau} \subseteq D$ be a submodule of $D$ such that $\mathrm{Im}{\sigma} \cong \mathrm{Im}{\tau}.$ Let $\kappa: B \rightarrow D$ and $\rho: \mathrm{Im}{\sigma} \rightarrow \mathrm{Im}{\tau}$ be $R$-module isomorphisms such that $\rho(p) = \kappa(p)$ for all $p \in \mathrm{Im}{\sigma}.$ Prove that if $n \in B,$ then $\kappa(n) \in D$ iff $n \in \mathrm{Im}{\sigma}$ AND prove that $\dfrac{B}{\mathrm{Im}{\sigma}} \cong \dfrac{D}{\mathrm{Im}{\tau}}.$
Proof of result: Note that if $n \in \mathrm{Im}{\sigma},$ then $\rho(n) \in \mathrm{Im}{\tau}.$ By hypothesis $\rho(p) = \kappa(p).$ So $\kappa(n) \in \mathrm{Im}{\tau}.$ Conversely, assume $\kappa(n) \in \mathrm{Im}{\tau}.$ Since $\rho$ is an isomorphism, it is surjective. So $\rho(p)=\kappa(n)$ for some $p \in \mathrm{Im}{\sigma}.$ By assumption, $\rho(p)=\kappa(p).$ So $\kappa(n)=\kappa(p).$ Since $\kappa$ is an isomorphism, $\kappa$ is injective. So $n=p \in \mathrm{Im}{\sigma}.$
Define $\varphi:B \rightarrow D/\mathrm{Im}{\tau}$ by $\varphi(n)=\overline{\kappa(n)}.$ Note that for $n_1,n_2 \in B,$ we have $\varphi(n_1n_2)=\overline{\kappa(n_1n_2)}=\overline{\kappa(n_1)\kappa(n_2)}=\overline{\kappa(n_1)} \cdot \overline{\kappa(n_2)}=\varphi(n_1)\varphi(n_2),$ because $\kappa$ is an isomorphism. Also, $\varphi(n_1+n_2)=\overline{\kappa(n_1+n_2)}=\overline{\kappa(n_1)}+\overline{\kappa(n_2)}=\varphi(n_1)+\varphi(n_2).$ So $\varphi$ is a module homomorphism. Now let $\bar{v} \in D/\mathrm{Im}{\tau}.$ Then $v \in D,$ so $v=\kappa(n)$ for some $n \in B$ because $\kappa$ is surjective. So $\varphi(n)=\overline{\kappa(n)}=\overline{v}.$ Note that $$\ker(\varphi)=\{n \in B: \overline{\kappa(n)}=\overline{0}\}=\{n \in B: \overline{\kappa(n)} \in \mathrm{Im}{\tau}\}\overset{\mathrm{(1)}}{=}\{n \in B: n \in \mathrm{Im}{\sigma}\}=\mathrm{Im}{\sigma}.$$ Thus by the First Isomorphism Theorem, the result follows.

Now back to the problem. All I need to show is that $\mathrm{Im}{\sigma} \cong \mathrm{Im}{\tau}$ and that $\rho(p) = \kappa(p)$ for all $p \in \mathrm{Im}{\sigma}.$
I have tried defining $$\rho= {\kappa^{-1}}\restriction_{\mathrm{Im}{\tau}} \circ \ \kappa \circ {\kappa}\restriction_{\mathrm{Im}{\sigma}} : \mathrm{Im}{\sigma} \rightarrow \mathrm{Im}{\tau}.$$
How do I show that $\rho$ is an isomorphism? Defining $\rho$ was difficult enough, now I have to deal with restrictions and compositions! 
Also, does $\rho(p) = \kappa(p)$ for all $p \in \mathrm{Im}{\sigma}$ follow directly once the isomorphism is defined?
Thanks for any help.
 A: You can use Five Lemma for diagram that results from assumptions: 
$\require{AMScd}$
\begin{CD}
    A @>\sigma>> B @>>>\dfrac{B}{\mathrm{Im}{\sigma}} @>>>0@>>>0\\
    @V \rho V V{\large\circlearrowleft} @VV \kappa V {\large\circlearrowleft} @VV{\text{induced by $\kappa$}}V @| @|\\
    C @>>\tau> D @>>>\dfrac{D}{\mathrm{Im}{\tau}}@>>>0@>>>0
    \end{CD}
A: Let $R = \mathbb{Z}$, $A = C = D = 0$, $B = \mathbb{Z}$. Then there is exactly one diagram between them und it is easily seen that your statement is wrong.
A: The diagram you have is
$$\require{AMScd}
\begin{CD}
A @>\rho>> C \\
@V\sigma VV @VV\tau V\\
B @>\kappa >> D
\end{CD}
$$
If $\rho$ and $\kappa$ are isomorphisms, it's obvious that $\def\coker{\operatorname{coker}}\coker\sigma\cong\coker\tau$:
for $b\in B$, define $f(b+\operatorname{Im}\sigma)=\kappa(b)+\operatorname{Im}\tau$ and verify


*

*$f$ is well defined

*$f$ is injective


Since you can consider also the diagram with $\rho^{-1}$ and $\kappa^{-1}$ you're done.
If $b=\sigma(a)$, then $\kappa(b)=\kappa(\sigma(a))=\tau(\rho(a))$ is in the image of $\tau$.
If $\kappa(b)=\tau(c)$, then $\sigma(\rho^{-1}(c)=\kappa^{-1}(\tau(c))$, so $b=\sigma(\rho^{-1}(c))$.

A proof with abstract computations. Consider the canonical projection $\pi\colon D\to D/\operatorname{Im}\tau$. Let $K=\ker(\pi\circ\kappa)$, so $B/K\cong D/\operatorname{Im}\tau$, because $\kappa$ is surjective.
Now prove that $K=\operatorname{Im}(\sigma)$.
If $b\in K$, then $\kappa(b)=\tau(c)$ for some $c\in C$, so $b=\sigma(\rho^{-1}(c))$.
If $b=\sigma(a)$ for some $a\in A$, then $\kappa(b)=\tau(\rho(a))$, so $b\in K$.
