While experimenting with orthogonal vectors I've noticed the following transformation: If $$ A = \begin{bmatrix}z & r \cr c & B\end{bmatrix} $$ is orthogonal, $z$ and $B$ square, and $I+z$ invertible, then the smaller matrix $$ A_1 = B-c(I+z)^{-1}r $$ is also orthogonal, it is not hard to show. I would like to know:

Reference request: Is this a known transformation? It must be, but I don't see it in books so far.

For the case where $z$ is $1\times1$, it is not hard to show that $\frac{1}{1+z}cr$ is bounded as $z$ tends to $-1$. But in fact, setting $A_1 = B$ in the case where $z=-1$, several numerical examples suggest that the mapping $O(n)\to O(n-1)$, $A\to A_1$, might even be continuous. It preserves inverses, but is not a homomorphism. The case $O(2)\to O(1)$ is constant on the two components and is continuous. So: is the mapping $O(n)\to O(n-1)$, $A\to A_1$, continuous?

Entirely speculative: if it is continuous, is $O(n)\to O(n-1)$ a bundle? At least the inverse image of each point seems to be $S^{n-1}$: Given $A_1\in O(n-1)$ and any unit $1\times n$ row vector $[z\ r]$ the matrix $$ A = \begin{bmatrix} z & r \cr -A_1r^T & A_1(I-\frac{1}{1+z}r^Tr) \end{bmatrix} \quad {\rm or,\ if\ } z=-1, \ \ A = \begin{bmatrix} -1 & 0 \cr 0 & A_1 \end{bmatrix}, $$ is in $O(n)$ and maps to $A_1$.