 Solution

If $m\in T_1$, then $rm=0$ for some nonzero $r$, andThus $\varphi (T_1)\subseteq T_2$. The formulais consequently independent of the representative and defines an $R$-module homomorphism on the quotient module.

If $\varphi$ is injective and $\Phi (m+T_1)=T_2$, then $\varphi (m)$ is torsion, so $r\varphi (m)=0$ for some nonzero $r$. Injectivity gives $rm=0$, hence $m\in T_1$. Therefore

The converse fails. For $R=\mathbb{Z}$, takeThen $\varphi$ is not injective, but quotienting $M_1$ by its torsion submodule leaves $\mathbb{Z}$, and $\Phi :\mathbb{Z}\to \mathbb{Z}$ is the identity.