Prove $C^{-1} (C\mathscr{R}) \subset \mathscr{R} + ker \; C$
Given that $C:\mathscr{X} \rightarrow \mathscr{Y}$ is a linear map where $\mathscr{X} and \; \mathscr{Y}$ are finite-dimensional linear vector spaces over the field $\Bbb{R}$.
$ker \; C := \{x \in \mathscr{X} \ | \ Cx=0 \}$
$\mathscr{R} \subset \mathscr{X}$
$C\mathscr{R} := \{y \in \mathscr{Y} \ | \ \exists x \in \mathscr{R}, \ y=Cx \}$
$\mathscr{P} \subset \mathscr{Y}$
$C^{-1} \mathscr{P} := \{x \in \mathscr{X} \ | \ Cx \in \mathscr{P} \}$
While I understand and can prove that $C^{-1} (C\mathscr{R}) \supset \mathscr{R}$, I do not fully understand where ker C comes from given that $\mathscr{R}$ is not necessarily a subspace of $\mathscr{X}$ (or if that even matters...)
Let $x \in C^{-1}(C \mathscr{R})$, then $Cx \in C \mathscr{R}.$ Hence there is $z \in \mathscr{R}$ such that $Cx=Cz.$ This gives that $u:=x-z \in ker C.$ Therefore
$$x=z+u \in \mathscr{R} + ker C.$$