It most books dealing with Cholesky decomposition, or it is variants, one finds a statement of the form if $A$ is symmetric $k\times k$ positive **semi-definite** (non-negative definite) then the $k\times k$ matrix $L$ solving 
$$
A=RR^{\top}.
$$
**Note:** I do not require that $A$ is positive *definite*, so $A^{-1}$ may not exist. However, I do require that it is symmetric.


Following his [post](https://mathoverflow.net/questions/225152/parametrization-of-positive-semidefinite-matrices), we see that under additional constraints there is a unique choice



> 
> **Theorem 10.9.** Let $A\in\mathbb R^{n\times n}$ be positive semidefinite of rank $r$. (a) There exists at least one upper triangular $R\in\mathbb R^{n\times n}$ with nonnegative diagonal elements such that $A = R^TR$. (b) There is a permutation $\Pi$ such that $\Pi^TA\Pi$ has a unique Cholesky factorization, which takes the form
>  $$
> \Pi^TA\Pi=R^TR,\quad R=\left(\begin{matrix} R\_{11} & R\_{12} \\ 0 & 0\end{matrix}\right),
> $$
>  where $R\_{11}$ is $r \times r$ upper triangular with positive diagonal elements.
> 
> 
> 


However, I cannot find the source of book or paper saying map $A \to R$ is continuous.



Reference request: continuity of Cholesky factor
A subtle issue is that $\Pi$ is not unique here. For instance, if


$$
A = \begin{bmatrix}
1 & 0 & 0\\\\
0 & 0 & 0\\\\
0 & 0 & 0
\end{bmatrix}
$$


then you can take both the identity and $(23)$ as the permutation. Similarly, if $A=I$, then any $\Pi$ will work (and $R=I$).


I don't think you can speak about continuity until you resolve this ambiguity.


