$V$ is finite-dim vector space, and $\text{U}$ its subspace, and $\text{V}'$ and $\text{U}'$ are their dual counterparts. $\text{U}^0 = \{ \phi \in \text{V}': \phi(u) = 0 $ for all u $ \in \text{U} \}$, then it's proved in Axler's 3.106 that:
$\text{dim U} + \text{dim U}^0 = \text{dim V}$
by taking $i \in \mathcal{L}(U,V) $, defined as $i(u) = i$ for $u \in \text{U}$ and, after applying the Fundamental Theorem of Linear Algebra on $i'$, showing that $\text{null } i' = \text{U}^0$ and $\text{range i}' = \text{U}$ (and $\text{dim V}' = \text {dim V}$)
What I don't understand, is how can the result for the single map $i$ be generalized to all the (sub)spaces. Can somebody clear that up for me? Thanks
It is incorrect to say that we have used a single map $i$ to conclude a statement about all subspaces. Rather, we have chosen a subspace $U$, constructed a map $i$ using that subspace $U$, and then used this map to make a conclusion about the subspace $U$. Because there was nothing special about the subspace $U$ (i.e. our choice of subspace was arbitrary), we were able to conclude a statement about all subspaces.
An equivalent perspective is this: what we have constructed is a map $i_U$ that depends on our choice of subspace $U$. By considering the relationship that every subspace $U$ has to the image and kernel of its corresponding map $i_U'$, we have reached a conclusion about all subspaces $U$.