Say we have a ring $R$ and let $A,B,C$ be $R$-mod. Then prove $\hom_R(N,-)$ is left exact (where $N$ is some fixed $R$-mod).
Basically we want to show that given that we know that $0\rightarrow A\stackrel{f}{\longrightarrow}B\stackrel{g}{\longrightarrow} C$ is exact, we want to show that their images under this functor satisfy the same property. That is, we want to show that $0\rightarrow \hom_R(N,A)\stackrel{\sigma}{\longrightarrow} \hom_R(N,B)\stackrel{\tau}{\longrightarrow} \hom_R(N,C)$is exact.
First of all I assume that the functor $\hom_R(N,-)$ takes the arrow $f$ in the category of $R$-modules to $\sigma$ (an arrow in the category of abelian groups), where for $\phi\in\hom_R(N,A)$, we have $\sigma(\phi)=f\circ\phi$. Similarly we define $\tau(\gamma)=g(\gamma)$ where $\gamma\in \hom_R(N,B)$. Then to prove that this is sequence is exact we have to prove two things. $\sigma$ is injective, and second $\text{im}(\sigma)=\ker(\tau)$.
The first part is easy since we have that if $\sigma(\phi)=0$, then for all $n$, we have that $f(\phi(n))=0$, but as $f$ is injective, we have that for all $n$, $\phi(n)=0$, so this means $\phi=0$, and thus $\sigma$ is injective.
Now let $\gamma\in \text{im}(\sigma)$. Then, there exists a $\phi\in \hom_R(N,A)$ such that $\sigma(\phi)=\gamma$. Then $\tau(\gamma)=g(\gamma)=g(\sigma(\phi))=g(f(\phi))=0$, as $g\circ f$ is $0$. To prove the other inclusion, let $\gamma\in \ker(\tau)$. Then $\tau(\gamma)=0$, which means $g(\gamma)=0$. Thus, for each $n\in N$ we have that $g(\gamma(n))=0$, so $\gamma(n)\in\ker(g)=\text{im}(f)$. Hence, for each $n\in N$ there exist an $a_n\in A$ such that $\gamma(n)=f(a_n)$. I was thinking about maybe defining $\phi(n)=a_n$, if we manage to show that $\phi\in \hom_R(N,A)$ then we would be done because $\gamma(n)=f(a_n)=f(\phi(n))=\sigma(\phi(n))$, so $\gamma=\sigma(\phi)$, but I am not sure at how to go on proving that $\phi\in \hom_R(N,A)$ since I am picking $a_n$ to be an arbitrary element element in $A$ that does $f(a_n)=\gamma(n)$.
Also, after I am done hammering that small detail, how would one go at proving that this functor is not right exact?
Thanks,