I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
Let $V$ be a vector space.
Let $V'$ be the dual space of $V$.
Let $W$ be a vector space.
Let $W'$ be the dual space of $W$.
Definition:
If $T\in\mathcal{L}(V,W)$, then the dual map of $T$ is the linear map $T'\in\mathcal{L}(W',V')$ defined by $T'(\varphi)=\varphi\circ T$ for $\varphi\in W'$.
Definition:
For $U\subset V$. The annihilator of $U$, denoted $U^0$, is defined by $$U^0=\{\varphi\in V' : \varphi(u)=0\text{ for all }u\in U\}.$$
The next result is from "Linear Algebra Done Right 3rd Edition". (on p.107)
3.108 $T$ surjective is equivalent to $T'$ injective
Suppose $V$ and $W$ are finite-dimensional and $T\in\mathcal{L}(V,W)$. Then $T$ is surjective if and only if $T'$ is injective.
Proof The map $T\in\mathcal{L}(V,W)$ is surjective if and only if $\operatorname{range} T=W$, which happens if and only if $(\operatorname{range} T)^0=\{0\}$, which happens if and only if $\operatorname{null} T'=\{0\}$ [by 3.107(a)], which happens if and only if $T'$ is injective.
3.107(a) is the following equation:
If $T\in\mathcal{L}(V,W)$, $$\operatorname{null} T'=(\operatorname{range} T)^0.$$
The author is very kind to the readers and, usually, there is no gaps in his proofs in this book.
But I felt a small gap in the proof of 3.108 above.
I felt the following fact is not so obvious.
Fact 1:
If $(\operatorname{range} T)^0=\{0\}$, then $\operatorname{range} T=W$.
Proof:
Assume that $(\operatorname{range} T)^0=\{0\}$ but $\operatorname{range} T\neq W$.
Then, $\dim \operatorname{range} T < \dim W$.
Let $v_1,\dots,v_k$ be a basis of $\operatorname{range} T$.
Let $v_1,\dots,v_k,\dots,v_l$ be a basis of $W$ ($k<l$).
Let $\varphi\in W'$ be a linear functional such that
$\varphi(v_i)=0$ for all $i\in\{1,\dots,l-1\}$ and
$\varphi(v_l)=1$.
Then, $\varphi\neq 0$ and $\varphi(v)=0$ for any $v\in\operatorname{range} T$.
So, $0\neq\varphi\in(\operatorname{range} T)^0$.
This is a contradiction.
Does the above fact immediately follow from some famous result?
By the way, the author commented about 3.107(a) as follows:
The proof of part (a) of the result below does not use the hypothesis that $V$ and $W$ are finite-dimensional.
And in 3.108, the author assumed that $V$ and $W$ are finite-dimensional.
So, we need to use the assumption that $V$ and $W$ are finite-dimensional in the proof of 3.108.
I guess we need to use the assumption that $W$ is finite-dimensional to prove Fact 1 above.
I guess we don't need the assumption that $V$ is finite-dimensional to prove 3.108.
