Cholesky: a homeomorphism from T_+ onto PD

Idea. The map $R\mapsto R^TR$ is polynomial; its inverse is constructed by composing already-known continuous maps.

Step 1: well-defined. For $R\in T_+$, $R$ is invertible, $R^TR$ is symmetric and $x^TR^TRx=|Rx|^2>0$ for $x\ne0$: $R^TR\in\mathrm{PD}$. The map is polynomial in the entries, hence continuous.

Step 2: injective. If $R^TR=R'^TR'$ with $R,R'\in T_+$, put $M=R'R^{-1}$, upper triangular with positive diagonal (products and inverses of such matrices are such). Then
$$M^TM=R^{-T}R'^TR'R^{-1}=R^{-T}R^TRR^{-1}=I,$$
so $M$ is orthogonal and upper triangular with positive diagonal; as in Problem 91, $M=I$, i.e. $R'=R$.

Step 3: surjective, with a continuous inverse. Let $A\in\mathrm{PD}$, $S=\sqrt A$ (positive definite, symmetric, hence in $GL_n$). By Problem 91, $S=QR$ with $Q$ orthogonal and $R\in T_+$. Then
$$A=S^2=S^TS=R^TQ^TQR=R^TR.$$
The assignment $A\mapsto R$ is the composite of $A\mapsto\sqrt A$ (continuous, Problem 89) and $S\mapsto R$ from the QR decomposition (continuous, Problem 91). It is a right and left inverse of $R\mapsto R^TR$ by the injectivity just proved.

Dimension count. $T_+\cong(0,\infty)^n\times\mathbb{R}^{n(n-1)/2}$, and $\dim\mathrm{Sym}_n=\frac{n(n+1)}2=n+\frac{n(n-1)}2$.

Numeric check. `numpy.linalg.cholesky` returns the lower factor $L=R^T$ with positive diagonal, and $LL^T=P$ held to $9\cdot10^{-16}$.

Remark. Cholesky's factorisation is the standard way to test positive definiteness and solve linear systems; the topological content here is that the factor depends *continuously* on the matrix, so it is stable under small perturbations. As a by-product, $\mathrm{PD}$ is homeomorphic to $\mathbb{R}^{n(n+1)/2}$.