link between completion of the universal enveloping algebra and an endomorphism of functor My question could be resume in the following way :

Let $\mathfrak{t} \to \mathrm{End}(V)$ a representation of an abelian Lie algebra into an infinite dimensional vector space. 
What can we say about this representation ? In particular, can we decompose it into a product of representation of dimension 1 ?

The question is probably a bit too general and I am just interested in the particular case bellow with $V = \widehat{\mathrm{U} \mathfrak{g}}$. 
Let $G$ be a complex algebraic group (of finite type and affine). Write $\mathfrak{g}$ its Lie algebra and choose a maximal torus $T \subseteq G$ of Lie algebra $\mathfrak{t}$. 
For example, I am particularly interested in $G = GL_n(\mathbb{C})$ but I think the following is general. In this case, $\mathfrak{g} = \mathfrak{gl}_n(\mathbb{C})$ the space of square matrices of size $n$, $T \simeq (\mathbb{C}^\ast)^n$ is the space of invertible diagonal matrices et $\mathfrak{t} \simeq \mathbb{C}^n$ is the space of diagonal matrices.
Then consider the adjoin representation $\mathrm{ad} : \mathfrak{t} \to \mathrm{End}( \mathfrak{g} )$. By the theory of representation, there exist a subspace $\mathcal{R} \subseteq \mathfrak{t}^\ast = \mathrm{Hom}_\mathbb{Z}( \mathfrak{t} , \mathbb{C} )$ thus than 
$$
\mathfrak{g} = \mathfrak{g}_0 \oplus \bigoplus_{\alpha \in \mathcal{R}} \mathfrak{g}_\alpha.
$$
where $\mathfrak{g}_0 = \mathfrak{t}$ and $\mathfrak{t}$ act on $\mathfrak{g}_\alpha$ by the character $\alpha$. 
Explicitly, we have 
$$
\mathfrak{g}_\alpha = \{ x\in \mathfrak{g} : \forall t\in \mathfrak{t} : \mathrm{ad}_t(x) = \alpha(t) \cdot x\},
$$
and $\mathfrak{g}_0 = \bigcap_{t\in \mathfrak{t}} \mathrm{Ker}( \mathrm{ad}_t )$.
Remark that a subspace of $\mathfrak{t}^\ast$ containing $\mathcal{R}$ is isomorphic to the space of characters. In fact, if $X^\ast(T) = \mathrm{Hom}_\mathbb{Z}(T, \mathbb{C}^\ast)$ is the set of characters, then it is isomorphic to the lattice in $\mathfrak{t}^\ast$ generated by differential of elements of $X^\ast(T)$. One can prove that this lattice contain $\mathcal{R}$.
If $\mathfrak{g}$ is a reductive Lie algebra, the space $\mathcal{R}$ is the root space and has more properties. 
(For a proof of the existence of the decomposition, every finite dimensional representation decompose into sum of irreducible representations but these last are of dimension $1$ because $\mathfrak{t}$ abelian so write thanks to a character $\mathrm{exp} \circ\alpha$.)
Write $\mathcal{W} = \mathbb{Z}[\mathcal{R}]$ the lattice generated by $\mathcal{R}$ in $\mathfrak{t}^\ast$ and consider the universal enveloping algebra $\mathrm{U} \mathfrak{g}$ of $\mathfrak{g}$.
We can then prove that we have a decomposition 
$$
\mathrm{U}\mathfrak{g} = \bigoplus_{\beta \in \mathcal{W}} \mathrm{U} \mathfrak{g}_\beta.
$$
This decomposition is compatible with the inclusion $\mathfrak{g} \hookrightarrow \mathrm{U}\mathfrak{g}$ and we have isomorphisms $\mathrm{U}\mathfrak{g}_\alpha \simeq \mathfrak{g}_\alpha$ for all $\alpha \in \mathcal{R} \cup \{0\}$. 
This is for example a consequence of the Poicare-Birkhoff-Witt theorem. 
Now there is my point :

consider the forgetful functor $F$ from the category $\mathrm{Rep}(G)$ of finite dimension representation of $G$ to the category of finite dimensional complex vector spaces et write $\widehat{\mathrm{U}\mathfrak{g}} = \mathrm{End}(F)$ the space of endomorphisms of the functor $F$. 

An element $x$ of $\widehat{\mathrm{U}\mathfrak{g}}$ is a collection $(x_r)_{r \in \mathrm{Rep}(G)}$ of elements $x_r \in \mathrm{End}(V_r)$ for all representation $r : G \to \mathrm{End}(V_r)$ compatibles with morphisms of representations.
We then can prove that we have two inclusion $G \hookrightarrow \widehat{\mathrm{U}\mathfrak{g}}$ et $\mathrm{U}\mathfrak{g} \hookrightarrow \widehat{\mathrm{U}\mathfrak{g}}$. (The existence of a natural morphism is pretty clear. To prove the inclusion, we use the fact that $G$ has a finite dimensional faithful representation and a sort of local analogue for $\mathrm{U}\mathfrak{g}$ with the representation into left-invariant differential operators).
We also have an action from $\mathfrak{t}$ onto $\widehat{\mathrm{U}\mathfrak{g}}$ witch implies a natural inclusion 
$$
\prod_{\beta \in \mathcal{W}} \widehat{\mathrm{U}\mathfrak{g}}_\beta  \hookrightarrow  \widehat{\mathrm{U}\mathfrak{g}},
$$
where 
$$
 \widehat{\mathrm{U}\mathfrak{g}}_\beta = \{ x \in \widehat{\mathrm{U}\mathfrak{g}} : \forall t\in \mathfrak{t} , \mathrm{ad}_t(x) = \rho(t) \cdot x \}.
$$

My question is :
  
  
*
  
*Is the previous morphism an isomorphism ? 
  
*Are the natural inclusions $\mathrm{U}\mathfrak{g}_\beta    \hookrightarrow  \widehat{\mathrm{U}\mathfrak{g}}_\beta$ isomorphisms ?
  

In this case, it would mean that the completion of the universal algebra $\mathrm{U}\mathfrak{g}$ is in fact equal to $\widehat{\mathrm{U}\mathfrak{g}} = \mathrm{End}(F)$. 
I have no idea if this result is easy to prove and/or it is known.
 A: It sounds like you want to prove that $\text{End}(F)$ is the profinite completion of the universal enveloping algebra $U(\mathfrak{g})$. I don't understand your strategy for proving this (in particular, the answer to your first question as stated is clearly "no," consider an infinite direct sum), but the following statement along those lines is true.

Theorem: Let $F$ be the forgetful functor from the category $\text{Rep}_f(\mathfrak{g})$ of finite-dimensional representations of a Lie algebra $\mathfrak{g}$ to finite-dimensional vector spaces. Then $\text{End}(F)$ is the profinite completion $\widehat{U(\mathfrak{g})}$ of $U(\mathfrak{g})$.

By the profinite completion of an algebra $A$ I mean the cofiltered limit over all finite-dimensional quotient algebras $A/I$. In fact the following more general claim holds.

Theorem: Let $A$ be an algebra and let $F$ be the forgetful functor from the category of $\text{Mod}_f(A)$ finite-dimensional $A$-modules to finite-dimensional vector spaces. Then $\text{End}(F)$ is the profinite completion $\widehat{A}$ of $A$. 

Sketch. The point is that $F$, while not representable, is pro-representable by the collection of finite-dimensional algebra quotients $A/I$ in the sense that
$$F(M) \cong \text{colim}_I \text{Hom}(A/I, M).$$
This is because $M$, being finite-dimensional, has the property that any morphism of $A$-modules $A \to M$ factors through a finite-dimensional quotient $A/J$ of $A$ as an $A$-module. The action of $A$ on $A/J$ in turn factors through a finite-dimensional quotient $A/I$ of $A$ as an algebra, and there is a natural $A$-module map $A/I \to A/J$ given by evaluation at $1$. Hence we can compute that
$$\text{Hom}(F, F) \cong \lim_I \text{Hom}(A/I, F) \cong \lim_I A/I$$
where in the second step we use the Yoneda lemma. $\Box$
For more amusing arguments along these lines see this blog post. 

This result does not hold if we replace $\text{Rep}_f(\mathfrak{g})$ by $\text{Rep}_f(G)$ for $G$ a connected Lie group with Lie algebra $\mathfrak{g}$ (unless of course $G$ is simply connected). Write $G \cong \widetilde{G}/Z$ where $\widetilde{G}$ is the universal cover. We have $\text{Rep}_f(\mathfrak{g}) \cong \text{Rep}_f(\widetilde{G})$, so the endomorphisms of the forgetful functor
$$F : \text{Rep}_f(\widetilde{G}) \to \text{Vect}_f$$
do correspond to $\widehat{U(\mathfrak{g})}$. On the other hand, the elements of $\widetilde{G}$ clearly act as automorphisms of the forgetful functor, and so we have a map 
$$\widetilde{G} \to \widehat{U(\mathfrak{g})}^{\times}$$
which is an inclusion as long as finite-dimensional representations of $\widetilde{G}$ separate points, which is true e.g. if it is compact. But $\text{Rep}_f(G)$ is the subcategory of $\text{Rep}_f(\widetilde{G})$ on which $Z \subset \widetilde{G}$ acts trivially, so the natural action of $\widehat{U(\mathfrak{g})}$ on the forgetful functor of $\text{Rep}_f(G)$ cannot be faithful, and must at least factor through the quotient obtained by setting every element of $Z$ to $1$. 
Explicitly, consider $G = SO(3), \widetilde{G} = SU(2)$. Here all of the relevant representation categories are semisimple. Write $V_{\lambda}$ for the irreducible representation of $SU(2)$ with highest weight $\lambda$, so that $\lambda \in \{ 0, 1, 2, \dots \}$. If $F_G$ denotes the forgetful functor $\text{Rep}_f(G) \to \text{Vect}_f$, then semisimplicity can be used to show that we have
$$\widehat{U(\mathfrak{sl}_2)} \cong \text{End}(F_{SU(2)}) \cong \prod_{\lambda \in \{ 0, 1, 2 \dots \}} \text{End}_{\mathbb{C}}(V_{\lambda})$$
while
$$\text{End}(F_{SO(3)}) \cong \prod_{\lambda \in \{ 0, 2, 4, \dots \}} \text{End}_{\mathbb{C}}(V_{\lambda}).$$
So the quotient operation performed by setting $-1 \in SU(2) \subset \widehat{U(\mathfrak{sl}_2)}^{\times}$ equal to $1$ is quite destructive: it kills half of the factors. 
