Following are items (e) and (f) of Exercise 11 page 179 of Hungerford's Algebra: [ the set T(A) is all torsion elements of A]
(e) If $0 \to A \to B \to C$ is an exact sequence of R-modules, then so is $0 \to T(A) \to T(B) \to T(C)$
(f) If $g :B \to C$ is an R-module epimorphism, then $g_T : T(B) \to T(C)$ need not be an epimorphism. [Hint: consider abelian groups.]
For (e) : $Ker(g)=Im(f)$. If $x \in Ker(g_T)$ then g_T(x)=0. Because $T(B) \subset$ B then $x \in Ker(g)=Im(f)$. Thus $x \in =Im(f_T)$. I cannot go the other direction because it seems not obvious how $x \in =Im(f)$ implies $x \in =Im(f_T)$ to proceed the rest.
For (f) : We need to find an element in T(C) which is not in $g_T(T(C))$. I don't know how and the hint seems not relevant?
This is exercise 11 on page 179 of Hungerford's Algebra book.
Let us recall the initial items that we are going to use.
From (a), we know that, if $A$ is a module over a commutative ring $R$, then $T(A) =\{a \in A | \text{ exists } r\in R, r \neq 0 \text{ and } ra=0 \}$.
From (b), we know that, if $R$ is an integral domain, then $T(A)$ is a submodule of $A$.
For item (d) to (f) $R$ is an integral domain.
From (d) we know that, if $f: A \to B$ is an $R$-module homomorphism, then $f(T(A)) \subseteq T(B)$; hence the restriction $f_T$ of $f$ to $T(A)$ ia an $R$-module homomorphism $T(A) \to T(B)$.
Now let us solve items (e) and (f).
Item (e): Suppose $0 \to A \xrightarrow{f} B \xrightarrow{g} C$ is an exact sequence.
Step 1: From item(d), we have that $f_T:T(A) \to T(B)$ and $g_T:T(B) \to T(C)$ are $R$-modules homomorphism.
Step 2: Since $0 \to A \xrightarrow{f} B \xrightarrow{g} C$ is an exact sequence, we have that $f$ is a monomorphism, so $f_T$ (the restriction of $f$ to the submodule $T(A)$) is also a monomorphism.
Step 3: Since $0 \to A \xrightarrow{f} B \xrightarrow{g} C$ is an exact sequence, we have that $\mathrm{Im}(f) = \mathrm{Ker}(g)$. Since $f_T$ is a restriction of $f$, we have $\mathrm{Im}(f_T) \subseteq \mathrm{Ker}(g)$, and since $\mathrm{Im}(f_T) \subseteq T(B)$. We have that $\mathrm{Im}(f_T) \subseteq \mathrm{Ker}(g) \cap T(B)= \mathrm{Ker}(g_T)$. So, we have proved that $\mathrm{Im}(f_T) \subseteq \mathrm{Ker}(g_T)$. Let us prove that the equality holds.
Given any $b \in \mathrm{Ker}(g_T)$. It means that $b \in T(B) \subseteq B$ such that $g_T(b) =0$. Since $g_T$ is a restriction of $g$, we have $g(b)=0$. So $b \in \mathrm{Ker}(g)$. Since $\mathrm{Im}(f) = \mathrm{Ker}(g)$, there is $a \in A$ such that $f(a)=b$. Since $b \in T(B)$, there is $r \in R$, $r\neq 0$ such that $rb=0$. So, $f(ra)= rf(a)=rb=0$. So $ra \in \mathrm{Ker}(f)$. But, $f$ is a monomorphism, so $ra=0$, so $a\in T(A)$. So, we have $f_T(a) =f(a)=b$. It means $b \in \mathrm{Im}(f_T)$. So, we have proved $\mathrm{Im}(f_T) \subseteq \mathrm{Ker}(g_T)$.
From steps 1, 2 and 3 above, we have that $0 \to T(A) \xrightarrow{f_T} T(B) \xrightarrow{g_T} T(C)$ is an exact sequence.
item (f): Consider $\Bbb Z$ and $\Bbb Z /4\Bbb Z$ as $\Bbb Z$-modules. Note that $\Bbb Z$ is an integral domain.
Let $g: \Bbb Z \to \Bbb Z /4\Bbb Z$ defined by $g(n) = n \bmod 4$. It is easy to prove that $g$ is a $\Bbb Z$-module epimorphism.
On the other hand, $T(\Bbb Z) =\{0\}$ and $T(\Bbb Z /4\Bbb Z) =\{\overline{0},\overline{2}\}$. So $g_T$ can not be an epimorphism.