On the natural isomorphism between $I$-torsion functor and direct limit of $\mathrm{Hom}$ functor Let $R$ be a commutative ring with unity with and let $I$ be a  proper ideal. (I'm not assuming $R$ is Noetherian.) For every $M \in R$-Mod, let  $\Gamma_I(M):=\{m \in M : I^n m=0$ for some $n\ge 1\}$.
If $f \in \mathrm{Hom}_R (M,N)$, it can be seen that $f(\Gamma_I(M)) \subseteq \Gamma_I(N)$, giving us a map $\Gamma_I (f):=f|_{\Gamma_I(M)} \in \mathrm{Hom}_R (\Gamma_I(M) , \Gamma_I(N))$. Thus we have a functor $\Gamma_I : R$-Mod $\to R$-Mod. 
How to show that $\Gamma_I$ is naturally isomorphic to the functor $\varinjlim \mathrm{Hom}_R(R/I^n, -)$ ? 
 A: $\require{AMScd}\newcommand\Hom{\operatorname{Hom}}\newcommand\Im{\operatorname{Im}}$For every $n\in\Bbb N$ we have an injective $R$-module homomorphism:
\begin{align}
&\tau_M^n:\Hom_R(R/I^n,M)\rightarrowtail\Gamma_I(M)&
&\xi\mapsto\xi(1+I^n)
\end{align}
Then for every $n\leq m$ we have a commutative diagram
\begin{CD}
\Hom_R(R/I^n,M)@>>>\Hom_R(R/I^m,M)\\
@V\tau^n_M VV@VV\tau^m_MV\\
\Gamma_I(M)@=\Gamma_I(M)
\end{CD}
hence we get an injective $R$-module homomorphism
$$\tau_M:\varinjlim_{n\in\Bbb N}\Hom_R(R/I^n,M)\to\Gamma_I(M)$$
Since
$$\Im\tau_M=\bigcup_{n\in\Bbb N}\Im\tau_M^n=\Gamma_I(M)$$
it follows that $\tau_M$ is, in fact, an $R$-module isomorphism.
Finally, for every $R$-module homomorphism, $\varphi:M\to N$, the commutative diagram
\begin{CD}
\varinjlim\Hom_R(R/I^n,M)@>>>\Hom_R(R/I^n,M)@>\tau_M^n>>\Gamma_I(M)\\
@VVV@VVV@VV\Gamma_I(\varphi)V\\
\varinjlim\Hom_R(R/I^n,N)@>>>\Hom_R(R/I^n,N)@>>\tau_N^n>\Gamma_I(N)\\
\end{CD}
proves the naturality of $\tau_M$ respect to $M$ giving rise to a natural isomorphism
$$\tau:\varinjlim\Hom_R(R/I^n,-)\to\Gamma_I$$
