I am working on the following problem and keep getting stuck.
Let $M$ an $R$-module and $A$ and $B$ submodules with $A\subseteq B$ and $\phi:M\rightarrow N$ an $R$-module homomorphism. Show that: $$ \phi(B)/\phi(A)\cong B/A\oplus(B\cap\ker\phi). $$
Here is what I have tried so far...
Assume the hypothesis. We know $\ker\phi$ is a submodule of $A$ and $\ker\phi$ is a submodule of $B$. Then by the 1st isomorphism theorem, we have $A/\ker\phi\cong \phi(A)$ and $B/\ker\phi\cong \phi(B)$. Therefore $\phi(B)/\phi(A)\cong (B/\ker\phi)/(A/\ker\phi)$.
Because $A\subseteq B$, we have $A$ to be a submodule of $B$ too. Notice $\ker\phi\subseteq A$. Then by the 2nd isomorphism theorem, we have $(B/\ker\phi)/(A/\ker\phi)\cong B/A$.
Claim $(B\cap \ker\phi)=0$. Recall $B=\{m\in M|m\in B\}$ and $\ker\phi=\{m\in M|\phi(m)=0\}$. Let $x\in(B\cap\ker\phi)$. Notice \begin{eqnarray} B\cap\ker\phi &=& \{m\in M|m\in B\}\cap \{m\in M|\phi(m)=0\}\\ &=& \{m\in M|m\in B\text{ and }\phi(m)=0\} \end{eqnarray} Then $x\in B$ and $\phi(x)=0$.
This is where I get stuck and am not sure how to proceed. I want to show that $x=0$ to prove $(B\cap \ker\phi)=0$. Any kind of help would be appreciated. I have no idea if I am on the right track or not. I assumed this problem was based off the isomorphism theorems, but the direct product is throwing me off a little bit. Is it an internal or external direct product?
Thanks for taking the time to help me.
Update after answer below:
Assume the hypothesis. Because $B$ and $A$ are submodules of $M$, we can consider $\phi:B\rightarrow N$ and $\phi:A\rightarrow N$. Notice $\phi$ is surjective when $\phi:B\rightarrow \phi(B)$ and $\phi:A\rightarrow \phi(A)$. Because $A$ and $B$ are submodules of $M$, we have $\phi(A)$ and $\phi(B)$ to be submodules of $N$. Additionally because $A\subseteq B$, we have $\phi(A)\subseteq \phi(B)$. Therefore $\phi(A)$ is a submodule of $\phi(B)$.
We have $\phi(B)/\phi(A)$ to be an $R$-module by a proposition. Then the natural projection map $$\varphi:\phi(B)\rightarrow \phi(B)/\phi(A)\text{ defined by }\varphi(\phi(b))=\phi(b)+\phi(A)$$ is an $R$-module homomorphism with $\ker(\varphi)=\phi(B)$. Therefore we have $$ B\xrightarrow[]{\phi}\phi(B)\xrightarrow[]{\varphi}\phi(B)/\phi(A) $$ Then $\varphi\circ\phi:B\rightarrow \phi(B)/\phi(A)$. By the 1st isomorphism theorem, we have $$B/\ker(\varphi\circ\phi)\cong \varphi(\phi(B)).$$ Because $\varphi\circ\phi$ is surjective by definition, we have $\varphi(\phi(B))=\phi(B)/\phi(A)$. Consider \begin{eqnarray*} \ker(\varphi\circ\phi) &=& \{b\in B|\varphi(\phi(b))=0\}\\ &=& \{b\in B|\phi(b)+\phi(A)=0\} \end{eqnarray*} where 0 is the zero coset, meaning $0=0+\phi(A)$. Then $\phi(b)=0$. Recall $\ker\phi=\{b\in B|\phi(b)=0\}$. This means $\phi(b)=0$ if $b\in B\cap \ker\phi$. Additionally $\phi(b)=0$ if $\phi(b)\in \phi(A)\subseteq \phi(B)$, which implies $b\in A\subseteq B$. Therefore \begin{eqnarray*} \ker(\varphi\circ\phi)&=&A+(B\cap\ker\phi)\\ &=& A\oplus (B\cap\ker\phi) \end{eqnarray*} Thus we have $$B/(A\oplus (B\cap\ker\phi))=B/\ker(\varphi\circ\phi)\cong \varphi(\phi(B))=\phi(B)/\phi(A).$$
Your statement makes no sense at all. The correct statement is:
The proof is very easy: We have a surjective map $$B \xrightarrow{\phi} \phi(B) \to \phi(B)/\phi(A),$$ where the second arrow is the canonical projection.
The kernel of this map is easily seen to be $A + (B \cap \operatorname{ker}\phi)$, which yields the desired result.