In the proof of showing $\operatorname{Ad}_{*} =\operatorname{ad} $ (adjoint representations of Lie Group(Algebra). I'am reading the Lee's Introduction to smooth manifolds, 2nd edition, p.534, Theorem 20.27 and stuck at understanding some equality :

Why the underlined equality is true? Here, $\operatorname{ad} : \mathfrak{g} \to \mathfrak{gl(g)}$ is the adjoint representation of $\mathfrak{g}$ and $\operatorname{Ad}_{*} :\mathfrak{g} \to \mathfrak{gl(g)}$ is the induced Lie algebra representation which is defined as follows (His book p.195) :

Here, the 'L' in the above underlined definition of $Y$ is defined as follows (His book, p.191) : Let $G$ be a Lie group. Let $v\in T_{e}G$ be arbitrary. Then define a (left invariant) vector field $v^{L}$ on $G$ by
$$ v^{L}|_g = d(L_g)_{e}(v)$$.
Note that by the proof of his book, Theorem 8.37 (p.191), the correspondence $L$ is an isomorphism $T_eG \to \operatorname{Lie}(G) =: \mathfrak{g}$ with inverse $\epsilon : \operatorname{Lie}(G) \to T_eG$, given by $\epsilon(X) := X_e$.
It seems that our question is somewhat basic but I don't know how to establish the equality rigorously. What should I notice?
First attempt : Let $H:=GL(\mathfrak{g})$ and let $\mathfrak{h}:=\mathfrak{gl(g)}$ be its Lie algebra. Let $\gamma : t \mapsto \operatorname{exp}tX$.
Then
$$\frac{d} {dt} |_{t=0} (\operatorname{Ad}(\operatorname{exp}tX)) 
= (\operatorname{Ad} \circ \gamma)^{'}(0) = d(\operatorname{Ad})_{\gamma(0)}(\gamma^{'}(0)) = d(\operatorname{Ad})_{e}(X_e) \in T_eH$$.
And,
$$\operatorname{Ad}_{*}(X) = (d(\operatorname{Ad})_{e}(X_e))^{L} \in \operatorname{Lie}H =: \mathfrak{h} = \mathfrak{gl(g)} $$
And why the difference appear?
I think that for the first equality, more correct form is,
$$(\operatorname{Ad}_{*}X)Y = (\frac{d} {dt} |_{t=0} (\operatorname{Ad}(\operatorname{exp}tX)))^{L}(Y) $$
Did I make point out well?
If so, it remains(the second equality) to show that
$$(\frac{d} {dt} |_{t=0} (\operatorname{Ad}(\operatorname{exp}tX)))^{L}(Y) = \frac{d} {dt} |_{t=0}(\operatorname{Ad}(\operatorname{exp}tX)Y) $$
How can we show this? How the passage of $Y$ into the derivative $\frac{d}{dt}|_{t=0}$  ( ; i.e., the second equality) possible? If we can use the Chain rule, how can we apply the chain rule more rigorously?
Can anyone helps?
 A: Here are my thoughts, which are too long to write down in the comment section. I believe what follows is substantially correct, with "holes"perhaps. I would appreciate any corrections/feedback.
Ad maps $G$ into $h=GL(\frak g)$ so Ad$_*$ maps $\frak g$ into $\frak h$ which is just $GL(\frak g)$ itself with the Lie bracket, and is written $\frak gl(g).$
Now, $X\in \frak g$ is identified with $X_e$ because of the $X\to X_e$ isomorphism of $\frak g$ with $T_eG.$ Similarly, Ad$_*X$ as an element of $\frak gl(g)$ is identified with an element of the tangent space at the identity in the tangent bundle of $\frak gl(g),$ which is just an element of $\frak gl(g)$ itself because the latter is a vector space. And so the formula Ad$_*X(Y)$ makes sense.
We also know that $X$ is a left-invariant vector field on $M$ so there is a one-parameter subgroup $\gamma$ which satisfies $X_{\gamma(t)}=\gamma'(t).$ And of course, $\gamma(t)=\exp tX.$
So, by definition of the pushforward, with which the induced Lie algebra representation is identified (I have not verified the validity of the identification), and with test function $f$, we get
Ad$_*X_ef=$Ad$_*X_{\gamma(0)}f=$Ad$_*\gamma'(0)f=\frac{d}{dt}|_{t=0}f($Ad$\circ \gamma(t))=\frac{d}{dt}|_{t=0}f($Ad$(\exp tX)).$
Hence, the first equality in the formula is correct.
Edit: here is a more careful proof:
$c^g:G\to G$ is defined by $c^g(x)=g^{-1}xg.$ Henceforth, we simply write this as $c.$
$\operatorname{Ad}:G\to \operatorname{GL}(\frak g)$ is defined by $\operatorname{Ad}(g)=c_*:\frak g \to \frak g,$ the induced Lie Algebra homomorphism.
$\operatorname{ad}:\frak g\to \operatorname{Hom}(\frak g, \frak g)=\frak gl(g)$ is defined by $\operatorname{ad}(X)=\operatorname{Ad_*}(X).$
$\operatorname{Hom}(\frak g, \frak g)$ is the Lie algebra of the Lie group $\operatorname{GL}(\frak g)$ in the same way as the Lie algebra $gl_n(\mathbb{C})$ of the general linear group $\operatorname{GL}_n(\mathbb C)$ can be identified with the space $M_n(\mathbb C)$ of $n×n$ complex matrices, with the Lie bracket $[A,B]=AB−BA$.
The key  identification is in what follows. By definition,
$\operatorname{Ad_*}(X)=(d\operatorname{Ad}_e(X_e))^L: \frak g\to \frak gl(g)$
which is the same as
$d\operatorname{Ad}_e(X):T_eG\to T_I(\operatorname{GL}(\frak g)).$
Hence,
$\operatorname{Ad_*}(X)=d\operatorname{Ad}_e(X)=\frac{d}{dt}|_{t=0}\operatorname{Ad}(\exp tX)dt.$
For the second part, it's just a matter of unwinding the definitions:
Fix $X\in \frak g.$ We have $\operatorname{Ad}\exp 0X=\operatorname{Ad}e=I$ (because $\operatorname{Ad}$ is a homomorphism.) Write $F:=\operatorname{Ad}\exp,$ take a $Y\in \frak g$ and compute, at $t=0$, the difference quotient:
$\tag1 \left(n\left(\left(F(t+\frac{1}{n}\right)X-I\right)\right)Y=n\left(\left(F(t+\frac{1}{n}\right)X)(Y)-Y\right).$
Now, define $F(tX)=\tilde F_X(t)$ and notice that $\tilde F_X(t)\in \operatorname{GL}(\frak g).$ Using these facts, rewrite $(1):$
$\tag2 n\left(\tilde F_X\left(t+\frac{1}{n}\right)-I\right)Y=n\left(\tilde F_X\left(t+\frac{1}{n}\right)Y-Y\right).$
The limit on the left-hand side is
$\tag3 \left(\frac{d}{dt}|_{t=0}\tilde F_X(t)\right)Y=\left(\frac{d}{dt}|_{t=0}\operatorname{Ad}\exp tX\right)Y$
so the limit on the right-hand side exists and is equal to it. Now notice that $\tilde F_X(0)Y=Y$ so the limit on the right-hand side of $(2)$ is
$\tag4 \frac{d}{dt}|_{t=0}\tilde F_X(t)Y=\frac{d}{dt}|_{t=0}\left(\operatorname{Ad}\exp X\right)Y$
