Let $G$ be a lie group and $\mathfrak{g}$ its lie algebra. Consider the category $Rep(G)$ of finite dimensional representations of $G$ and the category $Rep(\mathfrak{g})$ of finite dimensional representations of $\mathfrak{g}$. Morphisms in those categories are just $G$- resp. $\mathfrak{g}$ equivariant linear maps. There is an obvious functor $$d\colon Rep(G)\rightarrow Rep(\mathfrak{g}),\pi\mapsto d_e\pi$$ which maps a representation of $G$ to a representation of $\mathfrak{g}$ by taking the derivative at the neutral element. On morphisms, $d$ is just the identity because $G$ equivariant maps are also $\mathfrak{g}$ equivariant, what can be seen by putting 1-parameter subgroups of $G$ in the defining definition of beeing $G$ equivariant and taking the derivative.
If $G$ is simply connected, $d$ is bijective on objects, because Lie group homomoprhism are in bijection to the morphisms of their lie algebras, if the domain is simply connected.
Is $d$ bijective on morphisms too? So $d$ is an isomorphism of categories?
The question reduces to the question, whether every $\mathfrak{g}$ equivariant linear map between vector spaces is $G$ equivariant.
Best Answer
Let $V$ and $W$ be two representations of $G$ and $f : V \rightarrow W$ be invariant by $\frak{g}$, i.e. for all $X \in \frak{g}$ and $v \in V$: $$f(X.u) = X.f(u).$$ Let $X \in \frak{g}$ and $g=\exp(X)$. Define $$ \phi : \mathbb{R} \longrightarrow \mathrm{End}(V,W), \quad t \mapsto f \circ \exp(tX).$$ and $$ \psi : \mathbb{R} \longrightarrow \mathrm{End}(V,W), \quad t \mapsto \exp(tX) \circ f.$$ Then $$\frac{d}{dt} \phi(t) = f \circ X \circ \exp(tX) = X \circ f \circ \exp(tX) = X.\phi(t).$$ and $$\frac{d}{dt} \psi(t) = X \circ \exp(tX) \circ f = X.\psi(t).$$ Hence $\phi$ and $\psi$ both satisfy the differential equation. $$\begin{cases} \frac{d}{dt} y(t) &= X.y(t) \\ y(0) & = f \end{cases} $$ So $\phi=\psi$ by Cauchy-Lipschitz, and $f.g = g.f$.