Let $U,V$ be finite dimensional vector spaces over a field $\mathbb{F}$ and $T\in L(U,V)$. I want to prove that for any $v\in V$, the equation $T(u)=v$ has a solution for some $u\in U$ if and only if for any $v'\in V$ such that $T'(v')=0$, we have $\langle v,v'\rangle=0$.
I am having trouble seeing how I could show either direction of this statement. I know that if we have such a $v'\in V'$, then this $v'\in N(T')$, but I cannot see how this will imply that there is such a $u\in U$ that solves $T(u)=v$. Moreover, the other direction seems even more absurd to me. I would really appreciate any help here as I am stuck on how to prove this.
Note: Prime notation means it is the dual space or adjoint space if the prime is on a linear map.
Assume first that $T(u)=v$, and $f\in V':\,T'(f)=0$, then $$f(v)=\langle f,v\rangle=\langle f,\,T(u)\rangle=\langle T'(f),\,u\rangle=0 \, .$$ Note that this also implies $ f|_{R(T)}=0\implies T'(f)=0$ where $R(T)$ is the range of $T$.
For the other direction, if $v$ is not in the range of $T$, we need to find a $f\in V'$ such that $T'(f)=0$ but $f(v)\ne 0$.
Pick an arbitrary basis $b_1,\dots,b_r$ of $R(T)$, and extend it by $c_1=v,\,c_2,\dots,c_{n-r}$. Then define $f\in V'$ so that $$f(b_i)=0,\ f(c_1)=1,\ f(c_j)=0\ (j\ge 2)\,.$$