$u$ is onto if and only if $u^*$ is injective.
1. If $u^*(n)=0$ for some $n\neq0$, then for all $x\in\TT^d$: $1=\la x,u^*(n)\ra=\la u(x),n\ra$, which shows that $u$ cannot be onto.
2. If $u$ is not onto, then the commutative group $\TT^d/\im u$ is not trivial. Hence there is a character $\chi:\TT^d/\im u\rar S^1$. Denoting by $\pi:\TT^d\rar\TT^d/\im u$ the quotient map we get a character $\chi\circ\pi$ on $\TT^d$. Therefore there is some $n\in\Z^d\sm\{0\}$ such that $\chi\circ\pi=e_n$, i.e. for all $x\in\TT^d$: $e_n(u(x))=0$, which proves that $u^*(n)=0$.
As for the converse we need that every character of $\Z^d$ is of the form $n\mapsto e_n(x)$ for some $x\in\TT^d$. Actuall for every locally compact abelian group $G$ the set of characters of its dual group $\wh G$ is given by evaluating a character $\chi$ of $G$ at some point $x\in G$ - cf. e.g. Pontryagin Duality Theorem.
$u^*$ is onto if and only if $u$ is injective.
3. If $u^*$ is onto and $u(x)=0$, then for all $n\in\Z^d$: $1=\la u(x),n\ra=\la x,u^*(n)\ra$, i.e. for all $n\in\Z^d$: $1=\la x,n\ra$. Hence $x=0$.
4. If $u^*$ is not onto, then the commutative group $\Z^d/\im u^*$ is not trivial. Hence there is a non trivial character $\chi:\Z^d/\im u^*\rar S^1$. Denoting by $\pi:\Z^d\rar\Z^d/\im u^*$ the quotient map we get a non trivial character $\chi\circ\pi$ on $\Z^d$. Hence there is some $x\in\TT^d\sm\{0\}$ such that for all $n\in\Z^d$: $\chi\circ\pi(n)=e_n(x)$, i.e. for all $n\in\Z^d$: $1=e_{u^*(n)}(x)=\la x,u^*(n)\ra=\la u(x),n\ra$, which proves that $u(x)=0$.