$0\to A\to B\to C$ is exact if and only if $0\to \operatorname{Hom}_R(M,A)\to \operatorname{Hom}_R(M,B)\to \operatorname{Hom}_R(M,C)$ is exact

Question

I have to prove that if $R$ is a ring and $A,B,C$ are left $R$-modules then $0\to A\to B\to C$ is an exact sequence if and only if $0\to \operatorname{Hom}_R(M,A)\to \operatorname{Hom}_R(M,B)\to \operatorname{Hom}_R(M,C)$ is an exact sequence for every $R$-module $M$.

I have proved that if $0\to A\to B\to C$ is exact then $0\to \operatorname{Hom}_R(M,A)\to \operatorname{Hom}_R(M,B)\to \operatorname{Hom}_R(M,C)$ is exact. I want to prove the converse.
I have found in Proposition 4.1 here that the tensor product functor $N\to M\otimes _RN$ is right-exact, so my idea was to rewrite (or at least try to rewrite) said proof to prove that $N\to N\otimes _RM$ is left-exact, and then use the $\operatorname{Hom}-\otimes$ Adjunction Theorem to prove that if $0\to \operatorname{Hom}_R(M,A)\to \operatorname{Hom}_R(M,B)\to \operatorname{Hom}_R(M,C)$ is exact then $0\to A\to B\to C$ is exact.

Is this correct? Can we prove the converse in an easier way? Thank you for your time.

Answer 1

Let $$\tag{*}0\to N'\xrightarrow{\phi} N\xrightarrow{\psi} N'$$ be a sequence of morphisms such that $\psi\phi=0$ and whenever $M$ is a module the sequence $0\to\hom(M,N')\xrightarrow{\phi_*}\hom(M,N)\xrightarrow{\psi_*}\hom(M,N')$ is exact.

Let $K$ be the kernel of $\phi$ and let $i:K\to N'$ be the inclusion. The sequence $0\to\hom(K,N')\xrightarrow{\phi_*}\hom(K,N)\xrightarrow{\psi_*}\hom(K,N')$ is exact, and clearly $\phi_*(i)=\phi\circ i=0$, so that $i=0$. This tells us, of course, that $K=0$ and thus that $\phi$ is injective.

Let now $L$ be the kernel of $\psi$ and $j:L\to N$ the inclusion. The sequence $0\to\hom(L,N')\xrightarrow{\phi_*}\hom(L,N)\xrightarrow{\psi_*}\hom(L,N')$ is exact, and $\psi_*(j)=\psi\circ j=0$ because the image of $j$ is precisely the kenerl of $\psi$, and there is therefore a map $f:L\to N'$ such that $j=\phi_*(f)=\phi\circ f$. In particular, the image of $j$ is contained in the image of $\phi$. Since part of the hypothesis is that the image of $\phi$ is contained in the kernel of $\psi$ we actually have an equality.

We have shown that (*) is exact.