The following theorem is famous in module theory and is Proposition IV.$4.3$ of Hungerford's Algebra.
I) $A \overset{\theta}{\rightarrow} B \overset{\xi}{\rightarrow} C\rightarrow 0$ is an exact sequence of $R$-modules if and only if, for every $R$-module $D$, $$0 \rightarrow\text{Hom}_R(C,D) \overset{\bar{\xi}}{\rightarrow} \text{Hom}_R(B,D) \overset{\bar{\theta}}{\rightarrow} \text{Hom}_R(A,D) $$ is an exact sequence of abelian group. ($\bar{\xi}(f)=f\xi\ \ \forall f \in \text{Hom}_R(C,D)$ and $\bar{\theta}(g)=g\theta\ \ \forall g \in \text{Hom}_R(B,D)$)
I want to remove the fact that $\xi$ is an epimorphism and prove:
I') If $A \overset{\theta}{\rightarrow} B \overset{\xi}{\rightarrow} C$ is an exact sequence of $R$-modules, then for every $R$-module $D$ $$\text{Hom}_R(C,D) \overset{\bar{\xi}}{\rightarrow} \text{Hom}_R(B,D) \overset{\bar{\theta}}{\rightarrow} \text{Hom}_R(A,D)$$ is an exact sequence of abelian group.
Proof: $\xi \theta=0$ implies $\bar{\theta}\bar{\xi}=0$ and so $\text{Im}\bar{\xi} \subset\text{Ker}\bar{\theta}$. Hence, for all $f \in \text{Ker}\bar{\theta}$, there exists a unique homomorphism $\bar{f}: B/\text{Ker}\bar{\xi} \rightarrow D$ such that $\bar{f}(b+\text{Ker}\bar{\xi})=f(b)$ since $\text{Ker}\xi = \text{Im}\theta \subset \text{Ker}f $. Moreover, there exists an isomorphism $\phi: B/\text{Ker}\bar{\xi} \cong \text{Im}\xi$. Then $\bar{f}\phi^{-1}\pi:C \rightarrow D$ is an $R$-module homomorphism such that $\bar{\xi}(\bar{f}\phi^{-1}\pi)=f$, where $\pi :C \rightarrow \text{Im}\xi$ is a homorphism such that it is identity on $\text{Im}\xi$ and zero elsewhere. So $\text{Ker}\bar{\theta}\subset \text{Im}\bar{\xi} \ \ \blacksquare$.
If the proof is not correct, please tell me. Otherwise, let's continue.
There is another similar theorem for tensor product which is as follows (Proposition IV.$5.4$ in Hungerford's Algebra).
II) If $A \overset{f}{\rightarrow} B \overset{g}{\rightarrow} C \rightarrow 0$ is an exact sequence of left $R$-modules, then for every right $R$-module $D$ $$D \otimes_R A\overset{1_D \otimes f}{\rightarrow} D \otimes_R B \overset{1_D \otimes g}{\rightarrow} D \otimes_R C \rightarrow 0$$ is an exact sequence of abelian group.
Again, I want to remove the fact that $g$ is an epimorphism and prove:
II') If $A \overset{f}{\rightarrow} B \overset{g}{\rightarrow} C$ is an exact sequence of left $R$-modules, then for every right $R$-module $D$ $$D \otimes_R A\overset{1_D \otimes f}{\rightarrow} D \otimes_R B \overset{1_D \otimes g}{\rightarrow} D \otimes_R C$$ is an exact sequence of abelian group.
In this case, I face a problem.
Let $\pi: D \otimes_R B \rightarrow (D \otimes_R B)/ \text{Im}(1_D \otimes f)$ be the canonical epimorphims. If $g$ is an epimorphism, then we can show there exists an isomorphism $\alpha:\ (D \otimes_R B)/ \text{Im}(1_D \otimes f) \cong D \otimes_R C$ such that $\alpha (\pi(d \otimes b))=(1_D \otimes g)(d \otimes g)=d \otimes g(b)$. This is because there is a middle linear map $\beta:\ D \times C \rightarrow (D \otimes_R B)/ \text{Im}(1_D \otimes f)$ given by $(d,c) \mapsto \pi(d \otimes b)$, where $g(b)=c$, and this map is independent of the choice of $b$. Finally, $\alpha$ is an isomorphism implies $\text{Ker}(1_D \otimes g)=\text{Im}(1_D \otimes f)$.
In my case when $g$ is not epimorphism, I want to show $\alpha$ is a monomorphism to prove $\text{Ker}(1_D \otimes g)=\text{Im}(1_D \otimes f)$. However, I cannot find a bilinear map like $\beta$. Can anyone help me to prove the exact sequence when $g$ is not epimorphism? Is it possible?
The fact that I am able to prove (I') similar to (I), but I cannot prove (II') with the help of (II) because the proof of (I) and (II) have different natures is also interesting to me.