Let $A\in\mathbb R^{m \times k}$ be a matrix with full column rank. I want to show that there exist a neighborhood $\mathcal N$ of $A$ and a function $f:\mathcal N\to\mathbb R^{m\times (m-k)}$ such that
$f$ is continuous at $A$.
For all $B\in \mathcal N$, the columns of $f(B)$ form an orthonormal basis for $\text{col}(B)^\perp$, the orthogonal complement of the column space of $B$.
My attempt:
Let $M_A=I_m-A(A'A)^{-1}A'$. Then the columns of $M_A$ span $\text{col}(A)^\perp$. Since $\text{dim}\,(\text{col}(A)^\perp)=m-k$, we can select $m-k$ linearly independent columns from $M_A$, which then constitute a basis for $\text{col}(A)^\perp$. Let $j_1<\dots<j_{m-k}$ denote the indices of the selected columns.
For any $B\in \mathbb R^{m \times k}$, let $f(B)\in \mathbb R^{m\times (m-k)}$ be the matrix whose $r$th column corresponds to the $j_r$th column of $M_B=I_m-B(B'B)^{+}B'$, $r=1,\dots,m-k$, where $D^+$ denotes the Moore-Penrose inverse.
Then $f:\mathbb R^{m \times k}\to \mathbb R^{m\times (m-k)} $ is a function which is continuous at $A$ (by continuity of matrix inverse). Moreover, since $A$ and $f(A)$ both have full column rank, there exists a neighborhood $\mathcal N$ of $A$ such that $B$ and $f(B)$ both have full column rank for all $B\in\mathcal N$.
Hence $f$ satisfies points 1 and 2 above, except for the fact that the columns of $f(B)$ are not orthonormal. To fix this, we let $g:\mathcal N\to \mathbb R^{m\times (m-k)} $ be defined by
$$g(B)=f(B)[f(B)^\top f(B)]^{-1/2}$$
Then $g$ satisfies conditions 1 and 2 (continuity at $A$ follows from continuity of matrix square root and matrix inverse).
Is this proof correct? Thanks a lot for your help.