diffeomorphism from GL(n,$\Bbb {R}$) to SL($n,\Bbb {R}$) $\times \Bbb{R}^{\times}$

Question

It is Loring W.Tu "An introduction to manifolds" problem 15.9

$f$ has a inverse

$$g: (A, r)\mapsto AM_r$$

but how to show both $f$ and $g$ are $C^\infty$ ?

SL($n,\Bbb R$) have $n^2−1$ coordinates, not its entries

edit：from Chrystomath's answer, I understood that $f$ is $C^\infty$, but I cannot follow his proof about $g$, I'm just a beginner in manifold

Answer 1

$r\mapsto M_r$ is smooth: $$M_{r+h}=M_r+h\begin{pmatrix}1&0&\cdots&0\\0&0&\cdots\\\vdots\end{pmatrix}$$ so the map is differentiable, and the higher derivatives are $0$.

$(A,B)\mapsto AB$ is smooth: $$(A+H)(B+K)=AB+(AK+HB)+HK$$ so the derivative $(H,K)\mapsto AK+HB$ is linear, hence smooth (in finite dimensions).

$A\mapsto \det A$, $GL(n)\to\mathbb{R}^\times$ is multi-linear (from standard theory), hence also smooth.

Trivially, $x\mapsto 1/x$ for $x\ne0$ is smooth, and $A\mapsto(A,A)$ also.

Putting all these together gives that $f,g$ are smooth maps.