I need some help proving this statement. I don't really get why the set of functions that map every $v \in V$ to $0 \in W$ has to be empty. Why is this necessary for the map to be surjective? And what's the proof?
(I looked at this question: $\text{range}T=W$ if and only if $(\text{range}T)^{\circ}=\{0\}$, but I couldn't understand the given answer.)
Thanks in advance.
Well if $T$ is surjective, then $T(V)=W$ so $(range T)^0=W^0=\{v':v'(W)=\{0\}\}$ i.e. $(range T)^0$ is the set of all the linear functionals on $W$ which takes the entire space $W$ to $\{0\}$ which is obviously only the zero functional. So $(range T)^0=\{0\}.$
Suppose $(rangeT)^0=\{0\}$. If we denote $U=range T$, then $U$ is a subspace of $W$ and $U^0=\{0\}$ implies that the only linear functional that takes $U$ to $\{0\}$ is the zero functional which implies that $U=W$. (If $U$ were properly contained in $W$, then we could construct a nonzero functional that takes $U$ to $\{0\}$.) So $T$ is surjective.