= Solution
Here $\Phi$ acts on $\beta:V\to U$, so it is the same dualization construction with the source and target interchanged: $\Phi_{V,U}:L(V,U)\to L(U^*,V^*)$. This is the natural family interpretation of the symbol used in part (b).
If $\beta\alpha=0$, composition reversal gives $\alpha^*\beta^*=0$, so $\beta^*\in M^r(\alpha^*)$. Conversely, if $\delta\in M^r(\alpha^*)$, the <isomorphism> in part (b) writes $\delta=\beta^*$ uniquely. Then $\alpha^*\beta^*=0$ implies $(\beta\alpha)^*=0$; <injectivity> of dualization gives $\beta\alpha=0$. Thus
$$
\boxed{\Phi(M^l(\alpha))=M^r(\alpha^*)}.
$$
Similarly, $\alpha\beta=0$ is equivalent to $\beta^*\alpha^*=0$, giving
$$
\boxed{\Phi(M^r(\alpha))=M^l(\alpha^*)}.
$$
Equivalently, the reverse implications follow by taking double duals and using the canonical double-dual identification. The equalities are of subspaces under an <isomorphism>, not only inclusions.
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