Adjoint map $\text{Ad}:G\rightarrow \text{Aut}(\mathfrak{g})$ is smooth. Let $G$ be a lie group and let $\mathfrak{g}$ its lie algebra. Its not clear to me that $\text{Ad}:G\rightarrow \text{Aut}(\mathfrak{g})$ is smooth.
It is clear to me that if we have a matrix group any $g\in G$ gives that $\text{Ad}(g)$ is smooth. Because $Ad(g)Y=gYg^{-1}$ consists of rational polynomials in each entry. However this does not tell us that $Ad$ is smooth,  only that it sends each $g$ to a smooth.
How can we see that $Ad:G\rightarrow \text{Aut}(\mathfrak{g})$ is smooth? One approach might be to use that it is a group homomorphism between lie groups and then we need only show that it is continuous.

I do not think this question is a duplicate of (Adjoint action smooth (in Tapp's book)?) because it seems to me that this question on explains why each $\text{Ad}(g)$ is smooth, not $\text{Ad}$
 A: The direct way to prove this is as follows: Consider the map $F:G\times G\to G$ defined by $F(g,h):=ghg^{-1}$. This is obviously smooth, so its tangent map $TF:TG\times TG\to TG$ is smooth, too. Now we can restrict in the second component to to submanifold $T_eG=\mathfrak g$. On the other hand, the zero vector field $0:G\to TG$ is a smooth map. Hence we conclude that the map $G\times\mathfrak g\to TG$ defined by $(g,X)\mapsto TF(0(g),X)$ is smooth. But from the construction, it follows immedately that $TF(0(g),X)$ can be computed as $\tfrac{d}{dt}|_{t=0}gexp(tX)g^{-1}$. Hence we see that $TF(0(g),X)=Ad(g)(X)\in T_eG$, so we have actually proved that $(g,X)\mapsto Ad(g)(X)$ is smooth as a map $G\times\mathfrak g\to\mathfrak g$. But choosing a basis $\{X_i\}$ for $\mathfrak g$ with dual basis $\{\lambda_i\}$ for $\mathfrak g^*$ you get the entries of the matrix of $Ad(g)$ with respect to the basis $\{X_i\}$ as $a_{ij}=\lambda_i(Ad(g)(X_j))$. So these matrix entries depend smoothly on $g$, which exactly says that $Ad$ is smooth as a map $G\to GL(\mathfrak g)$.
