Rank of Lie group homomorphisms There is a theorem on page 84 of W. Boothby's An Introduction to Differentiable Manifolds and Riemannian Geometry, part of which reads as follows:
(6.14) Theorem If $F : G_{1} \to G_{2}$ is a homomorphism of Lie Groups, then the rank of $F$ is constant.
In the proof, we take $a \in G_{1}$ and $b = F(a)$ and then by the chain rule we have
$$DF(x) = DL_{b}(F(a^{-1}x)).DF(a^{-1}x).DL_{a}(x),$$
and it is briefly explained that since $L_{b}$ is a diffeomorphism and hence has a nonsingular Jacobian at each point, it follows that "the rank of $F$ is the same at $x$ and $a^{-1}x$, hence constant."
I cannot seem to get the point by this brief explanation, so I would appreciate any more explanations on this. Thanks.
 A: What the text is saying is that up to a modification by diffeomorphisms, the differential of a Lie group homomorphism is the same at every point. So in particular the rank of the map is constant.
The point is that homomorphisms essentially commute with left translations, i.e. for any $g\in G_1$,
$$F\circ L_g = L_{F(g)}\circ F.\qquad (1)$$
This can be seen directly from the homomorphism property, since for any $g,h\in G_1$,
$$F\circ L_g(h) = F(g\cdot h) = F(g)\cdot F(h) = L_{F(g)}\circ F(h).$$
Taking the differential of $(1)$ at the identity $e\in G_1$ gives
$$DF_g\circ DL_g = DL_{F(g)}\circ DF_e,$$
so we have the commutative diagram between the relevant tangent spaces
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
\begin{array}{111}
T_eG_1 & \ra{DF_e} & T_eG_2\\
\da{DL_{g}} & & \da{DL_{F(g)}}\\
T_gG_1 & \ra{DF_g} & T_{F(g)}G_2
\end{array}
$$
Since left translations are diffeomorphisms, the differentials $DL_{g}$ and $DL_{F(g)}$ are both linear bijections, so the rank of $DF_g$ for any point $g\in G_1$ is exactly the same as the rank of $DF_e$.
