Proving that two quotient modules are isomorphic

[kalish1]

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.

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**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.

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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.

This question has been sincerely crossposted on abstract algebra - Proving that two quotient modules are isomorphic - Mathematics Stack Exchange

 

[jvicens]

To show that $\rho$ is an isomorphism, we need to show that it is both injective and surjective.

For injectivity, let $p_1,p_2 \in \mathrm{Im}{\sigma}$ such that $\rho(p_1) = \rho(p_2)$. Then, by definition of $\rho$, we have $\kappa(p_1) = \kappa(p_2)$. Since $\kappa$ is an isomorphism, it is injective, so $p_1=p_2$. Thus, $\rho$ is injective.

For surjectivity, let $d \in \mathrm{Im}{\tau}$. Since $\kappa$ is surjective, there exists $b \in B$ such that $\kappa(b) = d$. Now, since $\rho = {\kappa^{-1}}\restriction_{\mathrm{Im}{\tau}} \circ \ \kappa \circ {\kappa}\restriction_{\mathrm{Im}{\sigma}}$, we have $\rho(\kappa\restriction_{\mathrm{Im}{\sigma}}(b)) = \kappa(b) = d$. Thus, $\rho$ is surjective.

Therefore, $\rho$ is an isomorphism.

Now, for the second part, we need to show that $\rho(p) = \kappa(p)$ for all $p \in \mathrm{Im}{\sigma}$. But this follows directly from the definition of $\rho$. For any $p \in \mathrm{Im}{\sigma}$, we have $\rho(p) = {\kappa^{-1}}\restriction_{\mathrm{Im}{\tau}} \circ \ \kappa \circ {\kappa}\restriction_{\mathrm{Im}{\sigma}}(p) = \kappa(p)$.

Thus, we have shown that $\mathrm{Im}{\sigma} \cong \mathrm{Im}{\tau}$ and that $\rho(p) = \kappa(p)$ for all $p \in \mathrm{Im}{\sigma}$. By the result proven earlier, we have $\dfrac{B}{\mathrm{Im}{\sigma}} \cong \dfrac{D}{\mathrm{Im}{\tau}}$.