Stalk of Lie groups morphism and Lie algebras

Question

Let $G,T$ be Lie groups and $\{U_i\}_i$ a system of neighboorhods of the unit of $G$. Assume that this system is final in the neighborhoods of the unit in the sense that for any open neighborhood $U\subset G$, there exists an $N$ such that for all $n\geq N$ such that $$U_n\subset U.$$ I'm trying to understand $$\text{colim}_i \text{Hom}(U_i,T).$$ We have a natural map $$\text{colim}_i \text{Hom}(U_i,T)\rightarrow \text{Hom}(\text{Lie}(G),\text{Lie}(T)).$$ Conversely, any Lie algebra homomorphism $\text{Lie}(G)\rightarrow \text{Lie}(T)$ gives rise to a Lie group morphism $U\rightarrow T$ for a sufficently small $e_G\in U\subset G$. By our assumption on $\{U_i\}_i$, this yields a natural map $\text{colim}_i \text{Hom}(U_i,T)\rightarrow \text{Hom}(\text{Lie}(G),\text{Lie}(T))$ and thus we would have $$\text{colim}_i \text{Hom}(U_i,T)\cong \text{Hom}(\text{Lie}(G),\text{Lie}(T)).$$ Is this reasoning correct? The reason why I'm interesded in this computation is that $\text{colim}_i \text{Hom}(U_i,T)$ can be seen as the stalk at $e_G$ of the sheaf $\text{Hom}(-,T)$.