The n-dual of a vector space

Studying linear algebra for my exam, I doubt arose. If I define the category $\mathcal{V}$ of all the vector spaces, and the functor $\mathcal{F}:\mathcal{V}\rightarrow \mathcal{V}$ given by $\mathcal{F}(A) = A^{*}$, where $A^{*}$ is the dual space of $A$, have any sense compose $\mathcal{F}\circ \mathcal{F}\circ ... \circ \mathcal{F}$ $n$ times? $\mathcal{F}$ have some particular propierty like differenciability or continuity? Sorry if the questions have some mistakes, I'm only an undergraduate student (a friend explain me that the "set" of all the vector spaces is a category, and the funtions between to categories are called functors).

Another question is, can I define a topology on $\mathcal{V}$, or a metric? If anyone can explain me or send me a link I'll really appreciate it.

• You need to worry about such things as defining a topology on a category (and also by ''saying that the ''set'' of all vector spaces is a category''). People worried about such things and realized that this thing we want to call a ''set'' is not a set in the set-theoretical sense, but rather a collection of objects which we call a ''class'' (and this has a well-defined mathematical meaning). Although you need to worry, you can pretty well survive in mathematics without clearing up this worry... – Patrick Da Silva Jun 27 '14 at 0:30
• Thanks for your answer! yes, I know that this is highly advanced, but the curiosity was superior hahaha. I study mathematics and my friends graduate always talks about categories, so I only want a small and easy answer, because I don't have enough knowledge. Thanks again! – El Peluca Sapeeee Jun 27 '14 at 0:45
• Molina : I think we've all been there! Don't be scared, just keep going ; I think that's the best advice. And asking questions doesn't hurt, here it's just fun :) – Patrick Da Silva Jun 27 '14 at 1:01

If $V$ is finite-dimensional, then $V^*$ is isomorphic to $V$, but not naturally-so, unless you have an inner-product or a non-degenerate bilinear form defined on $V$. On the other hand, $V^{**}$ is naturally-isomorphic to $V$ by the pairing of v**:=v*(v), i.e., a functional on $V^{**}$ evaluates an element of v* at v . If $V$ is infinite-dimensional, then $V$ embeds into $V^{**}$ (into its continuous dul, actually; please see below) but the embedding is not always an isomorphism (if it is, you say V is reflexive.).
I have never seen a way of deciding if this functor is continuous or differentiable; this is not done with functors, AFAIK, since categories do not often have a natural (or any at all) topology or norm. About the topology you can assign to $V^*$, there are many different topologies you can define on the dual :strong topology, weak topology, weak* topology, etc. You can maybe start with, e.g., http://en.wikipedia.org/wiki/Weak_topology and follow the links.