Zero-composition subspaces of a linear map ($M^l(\alpha )=\{\beta :\beta \alpha =0\},M^r(\alpha )=\{\beta :\alpha \beta =0\}$)

For $\alpha :U\to V$, maps in $M^l(\alpha )$ factor uniquely through $V/\mathrm{im}\alpha$; maps in $M^r(\alpha )$ have image in $\ker \alpha$. Thus these vector subspaces are naturally $L(V/\mathrm{im}\alpha ,U)$ and $L(V,\ker \alpha )$, of dimensions $(\dim V-\mathrm{rank}\alpha )\dim U$ and $(\dim U-\mathrm{rank}\alpha )\dim V$. Passing to a dual map reverses composition and interchanges these two spaces. This gives a basis-free route to equality of the ranks of a map and its dual.