SO(n) is a deformation retract of the general linear group

Idea. $GL_n^+\cong SO(n)\times\mathcal{P}_n$ where $\mathcal{P}_n$ is the (convex, hence contractible) cone of positive definite matrices.

Step 1: continuity of the polar decomposition. For $A\in GL_n^+$ put $P(A)=(A^TA)^{1/2}$ and $U(A)=AP(A)^{-1}$. The map $A\mapsto A^TA$ is continuous (polynomial), the positive square root is continuous on positive definite matrices (e.g. as the inverse of the continuous bijection $P\mapsto P^2$ of $\mathcal{P}_n$ which is a homeomorphism by the spectral theorem and the continuity of the eigen-projections; or by the holomorphic functional calculus $P^{1/2}=\frac1{2\pi i}\oint\sqrt z\,(z-P)^{-1}dz$), and the inverse is continuous. So
$$\Phi:GL_n^+\to SO(n)\times\mathcal{P}_n,\quad A\mapsto(U(A),P(A))$$
is continuous, with continuous inverse $(U,P)\mapsto UP$. (We showed $U(A)\in SO(n)$ when $\det A>0$ in Problem 8.) So $\Phi$ is a homeomorphism.

Step 2: the deformation. Define $H:GL_n^+\times[0,1]\to GL_n^+$,
$$H(A,t)=U(A)\big((1-t)P(A)+tI\big).$$
- $H$ is continuous (composition of continuous maps), and $(1-t)P+tI$ is positive definite so $H(A,t)\in GL_n^+$ (its determinant is $\det U(A)\det\big((1-t)P+tI\big)>0$).
- $H(A,0)=U(A)P(A)=A$.
- $H(A,1)=U(A)\in SO(n)$.
- If $A\in SO(n)$ then $P(A)=I$, $U(A)=A$ and $H(A,t)=A$ for all $t$.

Step 3: conclusion. $H$ is a deformation retraction of $GL_n^+$ onto $SO(n)$ fixing $SO(n)$ pointwise. So the inclusion $i:SO(n)\to GL_n^+$ and the retraction $r(A)=U(A)$ satisfy $r\circ i=\mathrm{id}$ and $i\circ r\simeq\mathrm{id}$ via $H$: they are inverse homotopy equivalences.

Remark. Consequently, all homotopy invariants of $GL_n^+(\mathbb{R})$ (fundamental group, homology) are those of $SO(n)$: for instance $\pi_1(GL_2^+)=\pi_1(S^1)=\mathbb{Z}$ (Problem 30), and $\pi_1(GL_n^+)=\mathbb{Z}/2$ for $n\ge3$.