Why is $\mathrm{range}(T^*)=(\mathrm{null}T)^0$? I'm reading Axler's Linear Algebra Done Right Third Edition and I'm struggling to figure out why exactly $\mathrm{range}(T^*)=(\mathrm{null}(T))^0$ by theorem 3.109 where $T\in L(V,W)$ and $V,W$ are finite dimensional. Later on this theorem is applied to prove the following result in exercise 36b):
Let finite-dimensional vector space $V$ and a subspace $U$ be given. Let $i\colon U\to V$ defined by $i(u)=u$. Prove that $\mathrm{range} (i^*)=U^*$.
If the above theorem is applied to this problem, we get: $\mathrm{range}(i^*)=(\mathrm{null}(i))^0=U^*$. But let's say we have $U\ne V$ and a subspace $W$ s.t. $W\oplus U = V$. Clearly, any $f\in W^*$ exist s.t. $f(u)=0$ and hence, $W^*\subset (\mathrm{null} i)^0$. But that makes no sense, since $g\in \mathrm{range}i^*$ only have $U$ as their domain ($i^*$ is defined by $i^*(\phi)=\phi \circ i$ for $\phi\in V^*$), whereas $f\in W^*$ have $V$ as their domain.
This brings me to my greater point of confusion. If $\mathrm{null}(i)\subset U\subset V$ (where $i$ is defined the same), why can't we say that $(\mathrm{null}(i))^0=V^*$?
Definitions:
$T^*(\phi)=\phi\circ T$ for $\phi\in W^*$ where $W^*$ is the dual space of $W$
$(null(T))^0$ is the annihilator of $null(T)$
 A: First, the main question: why is $\operatorname{range}(T^*) = (\operatorname{null}T)^0$?
First, let's note that in general (i.e. even if the spaces are infinite dimensional), we have $\operatorname{range}(T^*) \subset (\operatorname{null}T)^0$. Indeed, for any $g \in \operatorname{range}(T^*)$, there exists an $f \in W^* = \mathcal L(W,\Bbb C)$ such that $g = T^*(f) = f\circ T$. It follows that for any $x \in \operatorname{null}(T)$, we have
$$
g(x) = (f \circ T)(x) = f(T(x)) = f(0) = 0,
$$
which means that $g$ is an element of the annihilator of $\operatorname{null}(T)$.
To show that the spaces are equal, it suffices to note that they have the same dimension. To that end: suppose that $W$ is such that $V = \operatorname{null}(T) \oplus W$. For any $f \in \operatorname{null}(T)^0$, we have $f(v + w) = 0 + f(w)$. Thus, there is an isomorphism between $\operatorname{null}(T)^0$ and $W^*$, which has dimension
$$
\dim(W^*) = \dim(W) = \dim(V) - \dim\operatorname{null}(T).
$$
On the other hand, we also have
$$
\dim\operatorname{range}(T^*) = \dim\operatorname{range}(T) = \dim(V) - \dim\operatorname{null}(T).
$$

To your second question: the map $i$ is defined so that $i:U \to V$. It follows that $\operatorname{null}(i)^0$ is defined to be a subset of $U^*$.
To put this another way, given an element $f \in \operatorname{null}(i)^0$, it's not clear what $f(v)$ should be for an element $v \in V \setminus U$.
