Let $f \in \mathcal{L}(V, W)$. Moreover let's suppose $(e_1, ..., e_n)$ is a basis of $V$ and $f(e_i) = v_i$ where the $v_i$ aren't distinct (so there is at least $i \ne j$ such that $v_i = v_j$) so that $f$ is not one-to-one.
Then what I don't get is that the adjoint $f^* : W^* \to V^*$ is defined as : $f(v_i^*) = e_i^*$, but it doens't mean anything when $f$ is not one-to-one right ?
So why for example here they are talking about the linear transformation in the dual associated with $f$ which is $f$ but in the basis $B^*$ of $W^*$ and $C^*$ of $W^*$ ?
Thank you !
That is not how the adjoint map $f^*$ is defined. The definition is$$(\forall\alpha\in W^*):f^*(\alpha)=\alpha\circ f.$$
When we have a subset $\{f_1,\ldots,f_n\}$ of a vector space $V$, we define $f_1^*,\ldots,f_n^*\in V^*$ by$$f_i^*(f_j)=\begin{cases}1&\text{ if }i=j\\0&\text{ otherwise,}\end{cases}$$but this only makes sense if $\{f_1,\ldots,f_n\}$ is a basis. That's not the case with your $v_i$'s.