mod(R/I) is an exact abelian full subcategory of mod(R)

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This is a verification rather than a question.

So I was asked to show that if the category mod$(R/I)$ is an exact abelian full subcategory of category of mod$(R)$. For $R$ a commutative ring and $I$ an ideal of $R$.

Now $R \rightarrow R/I$ this canonical maps ensures that mod$(R/I)$ consists of only those $R$ modules which is annihilated by $I$.

It is abelian as kernels and cokernels are $R/I$ modules and $\textbf {Coimage}(f) \cong \textbf{Im}(f)$.

Now to prove it is full subcategory and the inclusion functor is exact doesn't it suffice to state my previously mentioned comment that mod$(R/I)$ consists of only those $R$ modules which is annihilated by $I$.

That means $$ \textbf {Hom}_{\textbf {mod(R/I)}}(M,N) = \textbf {Hom}_{\textbf {mod(R)}}(M,N)$$

Or is there some really explicit way to prove the full and exact portion?