Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 4 / 15G / a / Solution

{title2=$M^l(\alpha)=\{\beta:\beta\alpha=0\},\quad M^r(\alpha)=\{\beta:\alpha\beta=0\}$}

For $\alpha:U\to V$, maps in $M^l(\alpha)$ factor uniquely through $V/\operatorname{im}\alpha$; maps in $M^r(\alpha)$ have <image> in $\ker\alpha$. Thus these <vector subspaces> are naturally $L(V/\operatorname{im}\alpha,U)$ and $L(V,\ker\alpha)$, of dimensions $(\dim V-\operatorname{rank}\alpha)\dim U$ and $(\dim U-\operatorname{rank}\alpha)\dim V$. 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.

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 Incoming links: Zero-composition subspaces of a linear map