Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. For $X \in \mathfrak{g}$ then there is a smooth map $$ \Phi_X: \mathbb{R} \times G \rightarrow G $$ such that $\Phi_X(0,g) = g$ and $t \rightarrow \Phi_X(t,g)$ is an integral curves for $X$. Then the notes goes on to define $$ \Phi: \mathbb{R} \times G \times \mathfrak{g} \rightarrow \phi_X(t,g) $$ and states that this is a $C^{\infty}$ function.
What I don't understand is what is the topology defined on $\mathfrak{g}$? The proof simply considers a manifold $N$ in place of $\mathfrak{g}$ so I am wondering how do we view $\mathfrak{g}$ as a manifold? Thank you.
The Lie algebra $\mathfrak g$ is a finite-dimensional real vector space. Consider any norm on that vector space and consider the toplogy induced by that norm. Since all norms on a finite-dimensional real vector space are equivalent, you will always get the same topology.