$T:V \to W$ is surjective iff $T^*:W^*\to V^*$ is one-to-one Definitions:
$V^*$ is the dual space of vector space $V$ (The space of all linear functionals $T:V \to \mathbb R$).
For a given linear transformation $T:V \to W$, $T^*:W^* \to V^*$ denotes another linear transformation such that $\forall f \in W^*$, $T^*(f)= f \circ T$  .
Question:
Prove that $T:V \to W$ is surjective iff $T^*:W^*\to V^*$ is one-to-one.
My problem:  I can prove that if $T$ is surjective, then $\ker(T^*)=\{0_{w^*}\}$ and then conclude that $T^*$ is one-to-one.  But I can't prove the other side. What I mean is for a given $w \in W$ I can't find $v \in V$ such that $T(v)=w$.
 A: If $T$ is not surjective then $$\dim im T <\dim W$$ Take a $f\in W^*$ such that 
$$im T\subseteq \ker f.$$
Then $T^*(f)=0$.
A: More generally,
for $U$ a subspace of $W$,
define $\operatorname{ann}(U)=\{ f \in W^* : f(U)=0 \}$.
Then, $\ker T^* = \operatorname{ann}(\operatorname{im} T)$.
Note that $\operatorname{ann}(U)=0$ iff $U=W$. Indeed, if $U\ne W$, take $w \in W \setminus U$ part of a basis of $W$. Taking the coordinate of a vector with respect to $w$ defines a non-zero functional on $W$ that is zero on $U$. 
A: One can prove that if $T: V \to W$ is not surjective then $T^*: W^* \to V^*$ is not one-to-one.
Consider such a $T$. We know that $imT \neq W$, so there is a non-trivial linear subspace $U \subset W$ that is not reached by T. Consider $f \in W^*, f \neq 0$ such that $f|_{imT}: imT \to \mathbb{R}$ satisfies $f|_{imT} = 0$. (In finite dimension such a $f$ could be the inner product with a non-zero element of $U$).
Looking at the relation $T^*(f) = foT$, we can see that $f \in kerT^*$, and therefore $T^*$ is not one-to-one.
A: Note that a vector $v\in V$ is 0 if and only if for every element of the dual space, $f\in V^{*},$ we have that $f(v)=0.$ Then we have that $w=0$ if and only if $T^{*}w=0$ (injectivity of $T^{*}$), if and only if $\langle v,T^{*}w\rangle=0$ for every $v\in V,$ if and only if $\langle Tv,w\rangle=0$ for every $v\in V.$ But if $R(T)$ were a proper subspace of $W,$ then there would be nonzero elements of $W^{*}$ such that $\langle Tv,w\rangle=0$ for all $v\in V,$ so this says that $T$ is surjective.
To construct this nonzero element of $W^{*}$ explicitly, let $g\in W\setminus R(T),$ and let $S$ be the subspace of $W$ given by $R(T)\cup\mathrm{span}(\{g\}).$ On $S,$ define the linear functional $f(Tv+cg)=c,$ for all $v\in V$ and $c\in\mathbb{R}.$ Then if we set $C=\inf_{x\in R(T)}\|x+g\|$ (which is $>0$ since $R(T)$ is closed), we have that $|f(Tv+cg)|=|c|=(1/C)\inf_{x\in R(T)}\|cx+cg\|\leq (1/C)\|Tv+cg\|,$ and since $\|\cdot\|$ is sublinear on all of $W,$ we may extend this linear functional to all of $W$ by Hahn-Banach, yielding $\bar{f}\in W^{*}.$ Then $\|\bar{f}\|_{W^{*}}=1/C>0,$ so $\bar{f}\neq0,$ and $\langle Tv,\bar{f}\rangle=\bar{f}(Tv)=0$ for all $v\in V,$ as desired.
