I am trying to show that for any Lie group $G$ with $T_e G=:\mathfrak{g}$, $\exp:\mathfrak{g} \rightarrow G$ satisfies that $d_X \exp : T_X \mathfrak{g} \rightarrow T_{\exp(X)}G$ is invertible for any $X \in \mathfrak{g}$.
I know that $d_e \exp:\mathfrak{g} \rightarrow \mathfrak{g}$ is the identity map.
My attempt was to use this fact by rescaling so that we arrive in a neighborhood of the identity where $d\exp$ is invertible, and then use the fact that $\exp$ is a group homomorphism on $1$-parameter subgroups.
In detail:
We can assume there's a ball $B(0, \varepsilon) \subset \mathfrak{g}$ on which $d \exp$ is invertible, once we put a Euclidian metric on $\mathfrak{g}$.
Let $X \in \mathfrak{g} = M$ and take now $r := \frac{\varepsilon}{2 \lVert X \rVert}$. Then $\lVert rX \rVert = \frac{\varepsilon}{2}$, so $rX \in B(0, \varepsilon)$. We have:
$$ X = \left[ \frac{1}{r} \right] (r X) + \left\{\frac{1}{r}\right\} (rX) = k (rX) + f (rX) $$
with $k\in \mathbb N$ and $0 \leq f < 1$.
Since $\exp(tY)$ is the integral curve of the left invariant field generated by $Y$ and this integral curve is a $1$-parameter subgroup, we have $$ \exp(Y) = \exp(k(rY) + f(rY)) = \exp(rY)^k \exp(f rY) $$
Now take $V \in T_X \frak{g}$. An integral curve passing through $X$ with speed $V$ at time $t=0$ is $Y(t) = X + tV$.
We have: \begin{align*} d_X \exp(V) &= \left. \frac{d}{dt} \right\rvert_{t=0} \exp(Y(t)) \\ &=\left. \frac{d}{dt} \right\rvert_{t=0} \exp(rY(t))^k \exp(f rY(t)) \end{align*}
I don't know how to take it from here. I'm thinking I should use the fact that the rescaling maps $Y \mapsto rY$ and $Y \mapsto fY$ have invertible differentials, but I don't know what to do with the fact that we have a composition of multiplications in the RHS.
EDIT: I made some progress.
Setting $\phi_\lambda:\frak{g} \rightarrow \frak{g}$, $\phi_\lambda(Y) = \lambda Y$ we have: $$ \exp(Y) = (\exp \circ \phi_r)^k(Y) (\exp \circ \phi_{fr})(Y) $$ and \begin{align*} d_X \exp(V) &= \left. \frac{d}{dt} \right\rvert_{t=0} \exp(Y(t)) \\ &=\left. \frac{d}{dt} \right\rvert_{t=0} (\exp \circ \phi_r)^k(Y(t)) (\exp \circ \phi_{fr})(Y(t)) \end{align*}
Let $R$ denote right multiplication in $G$ and $L$ left multiplication. By the Leibniz rule (see https://math.stackexchange.com/questions/904231/differential-of-the-inversion-of-lie-group}) we have: \begin{align*} \left. \frac{d}{dt} \right\rvert_{t=0} & (\exp \circ \phi_r)^k(Y(t)) (\exp \circ \phi_{fr})(Y(t))\\ &= d_{(\exp \circ \phi_r)^k(Y(0))} R_{\exp \circ \phi_{fr}(Y(0))} \left( \left. \frac{d}{dt} \right\rvert_{t=0} (\exp \circ \phi_r)^k(Y(t)) \right) \\ &+ d_{(\exp \circ \phi_{fr})(Y(0))} L_{(\exp \circ \phi_r)^k(Y(0))} \left( \left. \frac{d}{dt} \right\rvert_{t=0} (\exp \circ \phi_{fr})(Y(t)) \right) \\ &= d_{(\exp \circ \phi_r)^k(X)} R_{\exp \circ \phi_{fr}(X)} \left( \left. \frac{d}{dt} \right\rvert_{t=0} (\exp \circ \phi_r)^k(Y(t)) \right) \\ &+ \left( d_{(\exp \circ \phi_{fr})(X)} L_{(\exp \circ \phi_r)^k(X)} \right) \circ \left( d_{\phi_{fr}(X)} \exp \right) \circ \left( d_X \phi_{fr} \right) (V) \end{align*}
The key point is that $d_{\phi_{fr}(X)} \exp = d_{fr X} \exp$ is invertible because $frX \in B(0, \varepsilon)$, so the whole composition of differentials in the second term is invertible, and by induction on $k$ so will be the other term. So I am left with a sum of invertible matrices. However, this is not always invertible. Is it invertible in this case?
