Product of exponential maps is local diffeomorphism

I want to solve the following problem (20-4 in Lee's book on smooth manifolds, second edition): I am quite new to these affairs, and would like you to check my proof.

My attempt at a solution:

In rigor, we should perhaps write $$F(X,Y)=\exp(X)|_A\exp|_B(Y)$$ From Prop. 20.8(e) of the same book, I know that the exponential map restricts to a diffeomorphism from a neighborhood of $0\in \mathfrak{g}$ to a neighborhood of $e\in G$. So there is a neighborhood $U$ of $0$ such that $\exp|_U$ is a diffeomorphism, and therefore both $\exp|_{U\cap A}$ and $\exp|_{U\cap B}$ are also diffeomorphisms.
Take $V=U\cap A\times U\cap B$. Notice that $(0,0)\in V$. Also, $$F|_V(X,Y)=\exp|_{U\cap A}(X) \exp|_{U\cap B}(Y)=\\ =\exp|_{U\cap A}(\pi_1(X,Y)) \exp|_{U\cap B}(\pi_2(X,Y))$$ So $F=(\exp|_{U\cap A}\circ\pi_1)\cdot (\exp|_{U\cap B}\circ\pi_2)$. Being the product of the composition of diffeomorphisms, $F$ is itself a diffeomorphism. $\square$

The differential of $exp$ at $0$ is the identity. This implies that $\partial_XF(X,Y)_{(0,0)}(u)=u$ and $\partial_YF(X,Y)_{(0,0)}(v)=v$. We deduce that $dF_{(0,0)}=Id$. The local inversion theorem implies that $F$ is a local diffeomorphism in a neighborhood of $(0,0)$.

For your argument $dim(U\cap A)<dim G$, so $exp_{\mid U\cap A}$ cannot be a diffeomorphism.

• But $\exp|_U$ is a diffeomorphism, right? The existence of such an $U$ is guaranteed by the proposition that I mentioned.
– Soap
May 14 '17 at 18:42
• Yes, it is a diffeomorphism May 14 '17 at 18:43
• Well, I thought that restrictions of diffeomorphisms were still diffeomorphisms
– Soap
May 14 '17 at 18:53
• restriction of local diffeomorphism to an open subset... May 14 '17 at 18:55
• $A$ (and therefore $U\cap A$) can be an open set...
– Soap
May 14 '17 at 21:50