I would like to prove the following fact: let $X$ a finitely generated $A-$module, $I$ is a filtered category, $F:I\to A-\text{mod}$ a functor. For each $i\in I,$ put $Fi=N_i,$ and $$\lim_{\to}{N_i}=(X,\{\rho_i: N_i\to N\}_{i\in I}).$$ Then $$\lim_{\to}\text{Hom}(M,N_i)\simeq \text{Hom}(M,N).$$ To simplify notations, I put $Y=\text{Hom}(M,N)$ and $$\lim_{\to}\text{Hom}(M,N_i)=(X,\{\phi_i: \text{Hom}(M,N_i)\to X\}_{i\in I})$$ Using the universal property of colimits with $Y$ and, for each $i\in I,$ the maps $$\begin{array}{rrcl}\psi_i:&\text{Hom}(M,N_i)&\to& Y\\ & \gamma&\mapsto &\rho_i\gamma\end{array}$$ I easily get the unique map $\Psi:X\to Y$ such that $\psi_i=\Psi\phi_i$ for each $i\in I.$ I would like to prove that $\Psi$ is an isomorphism but I don’t know how to go on. I think I have to exhibit the inverse of $\Psi$ but I’m not sure on how to use the fact that $I$ is filtered and $M$ is finitely generated.
My attemp: let $x_1,\dots x_n$ generators for $M.$ If $f\in\text{Hom}(M,N),$ then it is uniquely determined by $u_1=f(x_1),\dots,u_n=f(x_n)\in N.$
I know that $$N=\frac{\bigoplus N_i}{\sim}$$ where to element $n_i\in N_i$ and $n_j\in N_j$ are equivalent if there exist $k\in I$ and $\alpha:i\to k,$ $\beta:j\to k$ such that $F\alpha(n_i)=F\beta(n_j).$ Using this fact I can prove that for each $u_1,\dots u_n\in N$ there exist $i\in I$ and $v_1,\dots, v_n\in N_i$ such that $\rho_k(v_k)=u_k$ for each $k=1,\dots,n.$ Consequently I consider the map $$\begin{array}{rrcl}f_i:&M&\to& N_i\\ & x_k&\mapsto& v_k\end{array}$$ I think that a possible inverse is $$\begin{array}{rrcl}\Phi:&Y&\to& X\\ & f&\mapsto& \frac{\{f_i\}_{i\in I}}{\sim}\end{array}$$
This is my definition of filtered category: a category $I$ is filtered if satisfies the two following properties: $$\begin{array}{ll} (1)&\forall i,j\in I\ \exists x\in I, \alpha:i\to x,\ \beta:j\to x\\ (2)&\forall i,j,k\in I\ \forall \alpha:k\to i,\beta:k\to j\ \exists x\in I,\gamma:i\to k,\ \delta:j\to k:\ \gamma\alpha=\delta\beta. \end{array}$$
Is what I have written right, or are there some errors? If it is right is there a quick way to prove that $\Phi$ is the inverse of $\Psi?$ Or is there another way to prove this result quickly?