Suppose $A \in \mathcal M(n \times n; \mathbb R)$, $B \in \mathcal M(n \times m; \mathbb R)$ are fixed real matrices with $m < n$. Let $f : \mathcal M(m \times n; \mathbb R) \to \mathcal M(n \times n; \mathbb R)$ be defined by $X \mapsto A-BX$. I was alluded that this map is an open map if $B$ has full rank $m$.
I was thinking this can be showed by open mapping theorem. Since $X \mapsto BX$ is surjective and $\{BX: X \in \mathcal M(m \times n)\}$ is closed, then it is open by open mapping theorem. Translation is also open. This should be a flawed argument since then it does not matter whether $B$ has full rank. Could someone point me out where it is wrong and the correct argument?
I don't think that $f$ is open. First of all
$$g:\mathcal{M}(m,n)\to\mathcal{M}(n,n)$$ $$g(X)=BX$$
cannot be surjective because it is linear and the dimension of domain is $mn$ and the dimension of codomain is $n^2$ but you assume that $m<n$. Therefore the image of $g$ is a proper closed vector subspace of $\mathcal{M}(n,n)$.
So the image of $f=A-g$ is a proper closed (affine) subspace of $\mathcal{M}(n,n)$. Which obviously is never open.
But $f$ is open onto its image. Because $g$ is (as a surjective linear map onto its image). And in that case your argument is valid and indeed it does not depend on $B$.