About a Lie group We know that a real Lie group is a group that is also a finite-dimensional real smooth manifold, in which the group operations of multiplication and inversion are smooth maps.
And I saw an example of Lie group from Wikipedia:

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*The 2×2 real invertible matrices form a group under multiplication, denoted by $ \operatorname{GL}(2, \mathbf{R})$ :
$ \operatorname{GL}(2, \mathbf{R}) = \left\{A=\begin{pmatrix}a&b\\c&d\end{pmatrix}: \det A=ad-bc \ne 0\right\}.$
This is a four-dimensional noncompact real Lie group.
I want to ask how to prove the multiplication and inversion of matrices in this example is smooth?
 A: You have to see that it is a submanifold of $M(2,\mathbf{R})$ (which is a vector space so we know what smooth means for this one) and that multiplication is smooth in this one; for inversion you have to see how $\det$ is smooth on $M(2,\mathbf{R})$ and similarly for $A\mapsto Com(A)^T$; then the restrictions of all of these will be smooth on $GL(2,\mathbf{R})$ and you will get what you want
A: A map $f$ from a smooth manifold to $\Bbb R^n$  is smooth if and only if each of the component functions $f^i$ is smooth, i.e. by writing $f(p)=(f^1(p),f^2(p),\dots,f^n(p))$ for some real-valued functions $f^i$, you want to prove that $f^i$ are smooth.
The matrix multiplication is given by $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a'&b'\\c'&d'\end{pmatrix}=\begin{pmatrix}aa'+bc'&ab'+bd'\\a'c+c'd&cb'dd'\end{pmatrix}.$$ Each component function is just a polynomial, which is smooth.
In a similar manner, you can see that the component functions in matrix inversion are rational functions, which are again smooth.
