ok, first of all i'm not sure what you mean by the dual map but i can guess and it looks to me that it should be what you meant:

the dual $\displaystyle f': W^* \longrightarrow V^* $ of a linear map $\displaystyle f: V \longrightarrow W$ is defined by $\displaystyle f'(a)=af$ for all $\displaystyle a \in W^*.$

if that's the definition, then the answer to your question is "yes". to prove this, we only need to show that $\displaystyle f$ is onto because, since $\displaystyle f'$ is an isomorphism, we have

$\displaystyle \dim V = \dim V^*=\dim W^*=\dim W.$

so suppose it is not. then there exists a subspace $\displaystyle (0) \neq W_0$ of $\displaystyle W$ such that

$\displaystyle W=f(V) \oplus W_0.$

now let $\displaystyle 0 \neq a_0 \in W_0^*$ and define $\displaystyle a \in W^*$ by $\displaystyle a(w)=a_0(w_0), \ w \in W,$ where $\displaystyle w=f(v)+w_0.$ note that $\displaystyle a \neq 0$ and $\displaystyle a \in W^*$ is well-defined because the sum of $\displaystyle f(V)$ and $\displaystyle W_0$ is direct. clearly $\displaystyle a(f(v))=0$ for all $\displaystyle v \in V,$ i.e. $\displaystyle f'(a)=0,$ contradicting our hypothesis that $\displaystyle f'$ is an isomorphism.

the coverse is also true, i.e. if $\displaystyle f$ is an isomorphism, then $\displaystyle f'$ is an isomorphism too.