Suppose $L:V\to W$ is a linear transformation, and $L(M)\subseteq N$ for some subspaces $M\subseteq V$ and $N\subseteq W$.
A question I'm reading asks rather open-endedly if there is a relationship between $(L|_M)^*:N^*\to M^*$ and $L^*|_{N^0}:N^0\to M^0$? Here $N^*$ denotes the dual space of $N$, and $N^0$ the annihilator of $N$.
I'm not sure what relationship is being sought. Is there a known relationship between these two maps in the literature? Thanks.
For $\theta \in N^*$, $\forall x \in M, (L|_M)^*(\theta)[x]=\theta(L(x))\big(\in \mathbb{F}\big)$
For $\theta \in N^0$, $\forall x \in W, (L^*|_{N^0})(\theta)[x]=\theta(L(x)) \big(=0 $ if $x \in M$ since then $L(x) \in N$, so this map does indeed go from $N^0$ to $M^0$.$\big)$
In other words, the mappings do go from $N^*\rightarrow M^*$ and $N^0 \rightarrow M^0$ and they each map $\theta$ to $\theta\circ L$, so they behave in the same way.
It should be noted that in the finite dimensional case, if we take $w_1,\dots w_k$ to be a basis of $N$ and extend this to a basis $w_1,\dots w_n$ of $W$, then by taking the dual basis of $W^*$, $N^* \cong \langle v_1^*,\dots, v_k^*\rangle$ and $N^0=\langle v_{k+1}^*,\dots v_n^*\rangle$ so that $W^*=N^*\oplus N^0$. This can be viewed in much the same way as the decomposition of an inner product space into the direct sum of a subspace and it's orthogonal complement, and together with the previous relationship gives a way to think about "decomposing" $L^*$ on $W^*$.