Write each $B$ in $\mathbb{R}^{m\times n}$ in the form $\begin{bmatrix} B_{11} B_{12} \\ B_{21} B_{22} \end{bmatrix}$ such that $B_{11}$ is $k \times k$.
Let $U = \{B \in \mathbb{R}^{m \times n} : \det(B_{11}) \not= 0\}$, and let $F:U \to \mathbb{R}^{(m-k)\times (n-k)}$ be given by $$ F : B \mapsto B_{22} - B_{21} B_{11}^{-1} B_{12}. $$
The differential of $F$ is then a function $dF : U \to \mathbb{R}^{(m - k) \times(n - k)}$ (right?). Using the matrix cookbook, I have $$ dF(B) = dB_{22} - (dB_{21})B_{11}^{-1} B_{12} + B_{21}B_{11}^{-1} (dB_{11}) B_{11}^{-1} B_{12} - B_{21} B_{11}^{-1} (d B_{12}).$$
If $A$ is in $U$ and has rank $k$ (so $A_{11}$ is invertible and no larger submatrix of $A$ is invertible), how can I show that $dF(A)$ is surjective?
It is clearly false. Take $B_{2,2}=0$ and $B_{1,2}=0$