Let $Mod(R)$ be the category of modules over some ring $R$. We say $p: Mod(R) \to Mod(R)$ is a pre-radical if $p$ is a sub-functor of the identity functor over $Mod(R)$. Also, we say that a pre-radical is radical if $r(\frac{M}{r(M)})=\lbrace 0 \rbrace$ for every module $M \in Mod(R)$. Lets consider a radical $r$ and the following subclass of $Mod(R)$
$$\tau_{R}:= \lbrace M \in Mod(R) \: | \: r(M)=M \rbrace.$$
I want to prove this subclass is closed under extensions, that means that if I got an exact sequence
$$0 \to M' \to M \to M'' \to 0$$
where $M', M'' \in \tau_{R}$, then $M \in \tau_{R}$. I have read about this class and some books said this is a torsion class of a torsion theory this means this class is closed under extensions but nobody discuss the prove of this fact.
Let us assume for convenience of notation that $M'$ is actually a submodule of $M$ and $M''=M/M'$. Note that $M'=r(M')\subseteq r(M)$, so there is a surjective homomorphism $p:M''=M/M'\to M/r(M)$. Applying $r$ to this homomorphism, we get a commutative diagram
$$\require{AMScd} \begin{CD} r(M'') @>{}>> r(M/r(M))\\ @V{\cong}VV @V{0}VV \\ M'' @>{p}>> M/r(M) \end{CD}$$ where the left vertical map is an isomorphism by hypothesis and the right vertical map is $0$ since $r(M/r(M))=0$. This implies that $p=0$, but $p$ was surjective, so $M/r(M)=0$. That is, $r(M)=M$, as desired.