The polar-exponential decomposition of GL_n

Polar decomposition. For $A\in GL_n(\mathbb{R})$, $A^TA$ is positive definite. Let $P=(A^TA)^{1/2}$, the unique positive definite square root ($=\exp(\tfrac12\log(A^TA))$), and $U=AP^{-1}$. Then $U^TU=P^{-1}A^TAP^{-1}=I$, so $U\in O(n)$ and $A=UP$. Uniqueness: if $A=UP=U'P'$ then $A^TA=P^2=P'^2$ and uniqueness of the positive square root gives $P=P'$, hence $U=U'$. The map $A\mapsto(U,P)$ is continuous, since $A\mapsto A^TA$ is continuous and $P\mapsto P^{1/2}=\exp(\frac12\log P)$ is continuous (homeomorphism $\exp$ of the previous problem and its inverse). Conversely $(U,P)\mapsto UP$ is continuous. So $GL_n\cong O(n)\times\mathrm{PD}$.

Composition with $\exp$. The map $\Phi:(U,S)\mapsto Ue^S$ is the composition of $\mathrm{id}\times\exp:O(n)\times\mathrm{Sym}_n\to O(n)\times\mathrm{PD}$ (a homeomorphism) and the polar homeomorphism $O(n)\times\mathrm{PD}\to GL_n$. So $\Phi:O(n)\times\mathrm{Sym}_n\to GL_n(\mathbb{R})$ is a homeomorphism.

Positive determinant. $\det(Ue^S)=\det U\cdot e^{\operatorname{tr}S}$ has the sign of $\det U$. So $\Phi$ restricts to a homeomorphism $SO(n)\times\mathrm{Sym}_n\to GL_n^+(\mathbb{R})$ (and $O^-(n)\times\mathrm{Sym}_n\to GL_n^-(\mathbb{R})$).

Connectedness. $SO(n)$ is connected (it is $\exp(\mathrm{Skew}_n)$) and $\mathrm{Sym}_n$ is a vector space, so $GL_n^+(\mathbb{R})\cong SO(n)\times\mathrm{Sym}_n$ is connected. $GL_n(\mathbb{R})=GL_n^+\sqcup GL_n^-$, two open disjoint nonempty (for $n\ge1$) connected sets, so exactly two components.

Remark. Topologically, $GL_n^+(\mathbb{R})$ "is" $SO(n)$ times a contractible space: it deformation retracts onto $SO(n)$ (shrink $S\to0$). All the interesting topology of $GL_n^+$ is that of $SO(n)$.