Finding a natural transformation between the double dual functor and the trivial one My question is about to find a natural transformation from $t$ to the trivial functor, and $t$ goes from the following:
$$\mathbb{V} \to \mathbb{V^{*}} \to \mathbb{V}$$ 
then $t$ has to go from $\mathbb{V} \to \mathbb{V}$
where $\mathbb{V}$ is the category of vector spaces and $\mathbb{V^{*}}$ is that of the dual spaces.
I am really sorry if I am missing something but the thing is that I do not know how to begin or work with this, that is why I appreciate a very explicit answer with all the details, thanks in advance for your help, I really appreciate it :). 
I have found the following information, but I don't know if it helps, here http://en.wikibooks.org/wiki/Category_Theory/Natural_transformations where it says motivating example :)     
Here is the exact exercise :)
Show that exists a natural transformation from $t$ to the trivial functor, t is the one above 
 A: Let's assume your $t$ is the double dual functor $ V\mapsto V^{**}, f\mapsto f^{**}$. Then for every $V$ there's a map $\alpha_V:V\to V^{**}$ sending $v\mapsto (\xi\mapsto \xi(v))$. I claim this extends to a natural transformation $1_{\mathbb{V}}\to t$: if $f:V\to W$ then we need $\alpha_W\circ f=f^{**}\circ \alpha_V$. We have $\alpha_W\circ f(v)=(\zeta\mapsto \zeta(f(v)))$ and $f^{**}\circ\alpha_V(v)=f^{**}(\xi\mapsto \xi(v))=(\zeta\mapsto \zeta(f(v))$, as desired. 
Here it's worth spelling out $f^{**}$. $f^*:W^*\to V^*$ is given by $f^*(\zeta)(v)=\zeta(f(v))$. So $f^{**}:V^{**}\to W^{**}$ is given by $f^{**}(t)(\zeta)=t(f^*\zeta)$. In our example $t$ is the element $\xi\mapsto \xi(v)$ of $V^{**}$, and $f^{**}t(\zeta)=t(f^*\zeta)=t(v\mapsto \zeta(f(v)))=\zeta(f(v))$, as I claimed above. All of this is probably horribly messy to try to read, and I recommend that you try to write it out again yourself.
Anyway, we now have a natural transformation from the identity on finite-dimensional vector spaces to $t$. This goes in the opposite direction from what you wanted, but in the finite dimensional case $\alpha$ is a natural isomorphism, so take its inverse.
