Polar decomposition is a homeomorphism

Idea. Existence and uniqueness give a bijection; the formulas for the two directions are compositions of continuous maps (square root, inverse, product).

Step 1: uniqueness. If $A=UP$ with $U$ orthogonal and $P$ symmetric positive definite, then
$$A^TA=P^TU^TUP=P^2.$$
$P$ is a PSD matrix whose square is $A^TA$, so by uniqueness of the PSD square root $P=\sqrt{A^TA}$, and $U=AP^{-1}$.

Step 2: existence. For $A$ invertible, $A^TA$ is symmetric and positive definite ($x^TA^TAx=|Ax|^2>0$ for $x\ne0$). Put $P=\sqrt{A^TA}$: it is positive definite (its eigenvalues are the positive square roots), hence invertible. Put $U=AP^{-1}$. Then
$$U^TU=P^{-1}A^TAP^{-1}=P^{-1}P^2P^{-1}=I,$$
so $U\in O(n)$ and $A=UP$.

Step 3: continuity of $\Phi:A\mapsto(U,P)$. $A\mapsto A^TA$ is polynomial, $S\mapsto\sqrt S$ is continuous (Problem 89), $P\mapsto P^{-1}$ is continuous (Problem 68), and the product is continuous (Problem 67). So $P=\sqrt{A^TA}$ and $U=AP^{-1}$ are continuous functions of $A$.

Step 4: continuity of the inverse. $\Psi:(U,P)\mapsto UP$ is the restriction of the multiplication, continuous, and takes values in $GL_n$. By Steps 1-2, $\Psi\circ\Phi=\mathrm{id}$ and $\Phi\circ\Psi=\mathrm{id}$.

Consequences.
- $GL_n(\mathbb{R})\cong O(n)\times\mathrm{PD}$, and $\mathrm{PD}$ is a convex cone, hence contractible. So $GL_n$ deformation retracts onto $O(n)$ via $A\mapsto U$; likewise $GL_n^+\simeq SO(n)$.
- $O(n)$ has two components, so $GL_n$ has exactly two (Part 5).

Remark. The analogue for $n=1$ is $\mathbb{R}^\times=\{\pm1\}\times(0,\infty)$, the polar form $x=\operatorname{sgn}(x)|x|$ of a real number; for complex numbers it is $z=e^{i\theta}r$.