# To prove the Second Isomorphism Theorem for modules from the opposite direction:

Let $M$ be a R-module and let $S$ and $T$ be R-submodules of $M$. Then: $S/(S\cap T)\simeq (S+T)/T$

To prove the above Second Isomorphism Theorem for modules, we define the following:

Let $\phi$ be the mapping $S\rightarrow(S+T)/T$ such that $\phi(s)=s+T$.

Here, $\phi$ is a $R$-homomorphism with the kernel $\phi=S\cap T$.

By the first isomorphism theorem for modules, $S/(S\cap T)\simeq(S+T)/T$.

My question is whether it is possible to define the $R$-homomorphism from the opposite direction? Say,

Let $\psi$ be the mapping $(S+T)\rightarrow S/(S\cap T)$. If possible, how would such a map be defined in this case?

For element $x\in S+T$, you can write $x=s+t$ for $s\in S$ and $t\in T$, and then define $\psi(x)=s+S\cap T$. We have to check that this does not depends on the choice of $s$ and $t$, so if $x=s'+t'$ for $s'\in S$ and $t'\in T$, from $x=s+t=s'+t'$ we have $s-s'=t'-t$ which belongs to $S\cap T$. Hence $s+ S\cap T= s'+S\cap T$, and $\psi(x)$ does not depend on the choice of $s$ and $t$.
This is clearly a homomorphism. You can easily check that $\ker(\psi)=T$.