$\phi:M \rightarrow N$ is surjective, then $\hat{\phi}:M/Tor(M) \rightarrow N/Tor(N)$ is surjective

Question

Hello everyone, I already have done that $\phi(Tor(M)) \subset Tor(N)$. I'm stuck on the second part, so this is my attempt.

Since $\phi$ is an isomorphism, then since $Tor(M)$ and $Tor(N)$ are submodules of M and N respectively, it follows that $Tor(M)$ and $Tor(N)$ must be isomorphic as R-modules. So, if: $$\phi:M \rightarrow N$$ is a homomorphism of R-modules, then: $$\hat{\phi}:M/Tor(M) \rightarrow N/Tor(N)$$ must be a homomorphism of R-modules. Now, using the hint given in the answer, i have the following:

$\phi:M \rightarrow N$ is a surjective map and $\operatorname{Tor}(M)\subseteq M $ and $\operatorname{Tor}(N)\subseteq N $ and i proved that $\phi(\operatorname{Tor}(M)) \subseteq \operatorname{Tor}(N)$, so I only need to prove that $\hat{\phi}:M/\operatorname{Tor}(M) \rightarrow N/\operatorname{Tor}(N)$ is surjective. but stuck on this part. This is my attempt: $$\hat{\phi}(m+ \operatorname{Tor}(M)) = \hat{\phi}(m)+ \operatorname{Tor}(N)$$ Its surjective, since for $n+ \operatorname{Tor}(N) \in N/\operatorname{Tor}(N)$ we have:$$n+ \operatorname{Tor}(N) = \hat{\phi}^{-1}(m+ \operatorname{Tor}(M))$$ Then: $$n+ \operatorname{Tor}(N) = \hat{\phi}^{-1}(m)+ \operatorname{Tor}(M)$$

Answer 1

In fact, torsion has nothing to do with this. As a hint, you may want to show more generally that

if $f:M\to N$ is a surjective map and $M'\subseteq M$ and $N'\subseteq N$ are submodules such that $f(M')\subseteq N'$, then the induced map $f:M/M'\to N/N'$ is surjective.

Answer 2

You want to show every $\color{Red}{n}+{\rm Tor}(N)$ has a preimage $\color{Red}{m}+{\rm Tor}(M)$ under $\hat{\phi}$.

You already know every $\color{Blue}{n}$ has a preimage $\color{Blue}{m}$ under $\phi$. Any ideas?