Linear map between duals induced by linear maps between vector spaces Let $V, W$ be vector spaces over a field $F$ and let $\psi: V \to W$. Show that $\psi$ induces a linear map $\psi^{*}: W^{*} \to V^{*}$ naturally.
Although the question asks for a naturally induced linear map, it does not seem at all that easy to me. Any suggestions will be greatly appreciated!
 A: Hint:
$$\forall\,\,f\in W^*\;\;,\;\;\psi^*(f)(v):=f(\psi(v))$$
A: First, realize that the word 'natural' carries about as much meaning as the word 'obvious'. Don't let the word intimidate you.
Remember that $\psi^*:W^* \to V^*$, so if $w^* \in W^*$, then $\psi^*(w^*) \in V^*$.
Now select $v \in V$, and apply  $\psi^*(w^*) $, ie, make sense of $(\psi^*(w^*))(v)$, remembering that $\psi(v) \in W$.
The only way it 'fits' together is $w^*(\psi(v))$.
Rudin uses the suggestive notation $\langle v, \psi^*(w^*) \rangle$ for $(\psi^*(w^*))(v)$.
The above then reads $\langle v, \psi^*(w^*) \rangle = \langle \psi(v), w^* \rangle  $, which fits well with the Hilbert space notation.
A: DonAntonio's hint is good. Here is perhaps a more intuitive explanation:
We start with a given $\psi$, which tells us a specific road going from $V$ to $W$.
The elements of $W^*$ comprise all the different routes from $W$ to $F$.
Now, your goal is to come up with a general rule (called $\psi^*$) for taking routes from $W$ to $F$ and turning them into routes from $V$ to $F$.
Do you see how to use $\psi$ to accomplish this?
A: Here is a more pictorial hint:
You have $V\overset{\psi}{\to} W$ and then given $W\overset{f}{\to} F$ you want to find 
$V\overset{f^{\prime}}{\to} F$
Here $F$ is the ground field.  
