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, 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.

notwhat you are asking about here, so I have changed the title. Calling a Cholesky factor "square root" is slightly improper, although I have already heard it in various contexts. $\endgroup$Functions of matrices. It has a definition on how to extend an arbitrary scalar function to matrices (like you probably already studied with the exponential of a nonsymmetric matrix) in Chapter 1, including some remarks on branches and the principal square root, and then a chapter devoted to the properties of matrix square roots. $\endgroup$1more comment