A local diffeomorphism between Lie Algebra and Lie Group

Question

Let $\mathfrak{g}$ be a Lie Algebra and $G$ be the associated connected, simply connected Lie group. It is known that $\exp:\mathfrak{g}\rightarrow G$ is a local diffeomorphism.

Now let $\{e_1,\dots,e_n\}$ be a basis of $\mathfrak{g}$ and define $f:\mathfrak{g}\rightarrow G$ by $f(a_1e_1+\dots+a_ne_n)=\exp(a_1e_1)\dots\exp(a_ne_n)$. It is clamed that $f$ is a local diffeomorphism (for example in the proof of lemma 3, theorem 3.1). How do we prove this?

I thought of using the differential $df$, but I'm not sure how to calculate $df$ and how to conclude.

Answer 1

The domain of $f$ is a vector space, so you can essentially think of this $f$ as a parametrization and compute partial derivatives. But they said that $f$ is a local diffeomorphism only in a neighborhood of $0 \in \mathfrak{g}$. So $$\frac{\partial f}{\partial e_i}(0) = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} f(te_i) = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \exp(te_i) = e_i. $$Since $$\left\{\frac{\partial f}{\partial e_1}(0),\ldots, \frac{\partial f}{\partial e_n}(0) \right\} = \{e_1,\ldots, e_n \}$$is linearly independent, the Inverse Function Theorem says that $f$ is a local diffeomorphism in some neighborhood of $0$.

Answer 2

Left multiplication $L_g:G\to G$ is a local diffeomorphism for each $g \in G$, hence we have

$$ f(a^ie_i) \;\; =\;\; \prod_{i=1}^n L_{\exp(a^ie_i)} \;\; =\;\; L_{\exp(a^1e_1)} \circ \ldots \circ L_{\exp(a^ne_n)}. $$