$ \text{range } T' = (\ker T)^0$ Suppose that $V$ and $W$ are vector spaces. Denote $V'$ as the dual space of $V$.  Let $T \in \mathcal{L}(V,W)$. Denote the dual map of $T$ by $T' \in \mathcal{L}(W',V')$  such that
$$ T'(\phi) = \phi \circ T $$
and if $U$ is a subset of $V$, then $U^0$ is its annihilator.
If $V$ and $W$ are finite dimensional, then
$$ \text{range } T' = (\ker T)^0$$
Howerver, I do not know if this is still true if one of the vector spaces is infinite dimensional. In the Sheldon Axler's Linear Algebra Done Right (3rd ed.), he proves that $\text{range } T' \subset (\ker T)^0$ without the hypothesis of their being finite dimensional. However, he uses this hypotheses to prove the inclusion in the other direction. So my question is: does the equality still holds if one of the vector spaces is infinite dimensional? If does not, could you show a counterexample.
Just consider during the explanation that my level of linear algebra is roughly that of the book Linear Algebra Done Right. So if the explanation requires more advanced knowledge, just explain it briefly, just to give me some feeling.
Edit:
$V'$ is the algebraic dual:
$$V' = \mathcal{L}(V,\mathbb{F}) $$
where $\mathbb{F}$ is the field, that is, $V$ is over the field $\mathbb{F}$.
 A: Yes, this does hold. Let $T : V \to W$ be linear.
One thing you can always do in vector spaces is complement subspaces. That is, given a subspace $X \le V$, you can find $Y \le V$ such that $X \oplus Y = V$. In finite dimensions this is done with bases, but in infinite dimensions, it's done with some kind of axiom of choice argument.
Aside: In infinite dimensions, it is common to attach a norm or topology to the space, forming a normed or topological vector space; most applications of infinite-dimensional vector spaces have natural topologies or norms, and yield more useful results. There are separate definitions for dual, adjoint, and complemented for spaces with norms or topologies. What we are discussing is the lesser-used algebraic versions of these terms. I mention this because it's a hard problem figuring out when a subspace can be topologically complemented, whereas algebraic complements, as I said, are guaranteed.
Let's start by finding complements for $\operatorname{ker} T$ in $V$ and $\operatorname{im} T$ in $W$; call them $V_0$ and $W_0$ respectively. Let
$$S : V_0 \to \operatorname{im} T : v \mapsto Tv,$$
which is to say, restricting $T$'s domain and codomain. I claim that $S$ is bijective and hence invertible.
To show $S$ is injective, it still suffices to show $\operatorname{ker} S \subseteq \{0\}$. Suppose $Sv = 0$. Then $Tv = 0$, and $v \in V_0$. But this means that
$$v \in V_0 \cap \operatorname{ker} T = \{0\} \implies v = 0$$
as needed, since $V_0$ and $\operatorname{ker} T$ sum directly. Thus, $S$ is injective.
Now, let's show that $S$ is surjective. Start with $w \in \operatorname{im} T$. Then, we know there exists some $v \in V$ such that $Tv = w$. As $V = \operatorname{ker} T \oplus V_0$, there exist $v_1 \in \operatorname{ker} T$ and $v_2 \in V_0$ such that $v = v_1 + v_2$. Hence,
$$w = Tv = T(v_1 + v_2) = Tv_1 + Tv_2 = 0 + Tv_2 = Sv_2,$$
since $v_1 \in \operatorname{ker} T$. Thus $S$ is surjective.
So, how does this help us? Consider a $\psi \in (\operatorname{ker} T)^0$. We wish to show there exists some $\phi \in W'$ such that $\phi \circ T = \psi$. Define $\phi$ by defining it separately on $\operatorname{im} T$ and on $W_0$; since these space sum directly to $W$, we are free to do this without fear of contradiction.
For $w \in \operatorname{im} T$, let $\phi(w) = \psi(S^{-1}w)$. For $w_0 \in W_0$, let $\phi(w_0) = 0$ (although, any linear function could have been chosen instead of $0$ here!).
Now, let's show that $T'(\phi) = \psi$. Given $v \in V$, we have
$$(T'(\phi))(v) = \phi(Tv) = \psi(S^{-1}Tv)$$
Since $v \in V$, we can write $v = v_1 + v_2$, where $v_1 \in \operatorname{ker} T$ and $v_2 \in V_0$, so
$$\psi(S^{-1}Tv) = \psi(S^{-1}T(v_1 + v_2)) = \psi(S^{-1}Tv_2)$$
But, as $v_2 \in V_0$, we have $Tv_2 = Sv_2$, hence
$$(T'(\phi))(v) = \psi(S^{-1}Sv_2) = \psi(v_2).$$
However! Recall that $\psi \in (\operatorname{ker} T)^0$, so $\psi(v_1) = 0$. Thus,
$$(T'(\phi))(v) = \psi(v_2) + 0 = \psi(v_2) + \psi(v_1) = \psi(v_1 + v_2) = \psi(v),$$
proving $T'(\phi) = \psi$. Thus, the equality between sets holds in general.
A: I'm assuming by $V'$ you mean the algebraic dual, i.e., all $\mathbb{F}$-linear maps $V\to\mathbb{F}$.  If $V$ is a topological vector spaces over a topological field $\mathbb{F}$, then $V'$ might stand for the continuous dual.  I'll use the notation $V^\vee$ for the algebraic dual of $V$.
Given any element $\phi\in(\ker T)^0$, define $\psi\in(T(V))^\vee$ by $\psi(T(v)) = \phi(v)$.  This is well-defined, so we have a linear map $\psi\colon T(V)\to\mathbb{F}$.  If we can extend this to $\tilde\psi\colon W\to\mathbb{F}$, then we are done.
If we assume the axiom of choice, then we can extend a Hamel basis of $T(V)$ to a Hamel basis of $W$ and hence define $\tilde\psi$ to be zero on the additional basis elements.
Addendum: The statement "on every vector space over $\mathbb{F}$ and any subspace, we can extend every linear functionals on the subspace to the full space" is apparently equivalent to the statement "there exists nontrivial linear functional on every nontrivial $\mathbb{F}$-vector space" arxiv:1901.05146.  To the best of my knowledge, it is still open whether this actually implies the axiom of choice or some weaker variants.
A: The equality between the annihilator of the kernel of an operator and the range of its dual does not hold in infinite dimensional spaces in general, because while the annihilator of the kernel is always a closed subspace, the range of an operator need not be. An explicit counter-example can be found here.
