\section{Some category theory}
Why do we do this?

Firstly, the dual is a (contravarient) functor: if you want to
understand it, you need to understand functors.

Second, the fact that the double-dual of a finite-dimensional vector
space is ``naturally'' isomorphic to the original vector space
is a statement about natural transforms of functors, so you'll need to
learn that, too.

Thirdly, linear algebra without category theory is confusing: all the
category theory is there, lurking, and if you try to ignore it, you'll
suffer. We may as well make it explicit.

Fourthly, category theory is a beautiful and fundamental part of
mathematics, showing common algebraic underpinnings of widely
disparate areas, and providing important insight and unification.

Lastly, linear algebra is a showcase of category theory -- many of the
important concepts are here, and it's an easy enough context that you
can do a \emph{lot} of math. I like to say that with easy objects, we
can build up complex structures: what's interesting are the
connections and interrelations of the objects: with complicated objects,
even understanding the objects is difficult and interesting.

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lotta category theory
use EGs from set theory

(egs of "is this a category?")

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functors; e.g.s from set theory
(esp. power set -- both co-varient and contravarient)

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when Hom-sets have structure

the only structure one assumes by default on Hom-sets is a set.
sometimes there's more
(and what we requires is that composition is a good map)
(remember, we can put whatever structure we want on the objects,
 but category theory only sees the maps)

It's particularly interesting when Hom-sets have the same structure
as the category itself
(okay, it's -very- interesting: you can use the subject to study itself more)
(egs: sets, top, algGeo, and of course linear algebra)

in our case, Hom(V,W) has the structure of a vector space,
and composition is bilinear! (NB: not linear)

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representable functors:

the identity functor is represented in Sets and VectSpace

exercises:
show that power set contravarient functor is represented by
$\set{0,1}$;
  (call it subobject classifier; get topoi)
give a counting argument with finite sets to show that the
power set covarient functor cannot be represented.

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(sidebar: words and pictures:
commutative diagrams are as rigorous as words;
people use words b/c they're more familiar,
more linear,
and easier to typeset.
But (mathematical) reality isn't words -- it just -is-,
and words, diagrams -- these are just attempts to capture it.

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Hom as a bifunctor
(cute diagrams -- spend a bit of time)

%--------------------
``Naturality''
discuss natural transforms of functors
(note that clearly dual -cannot- be natural, as it's contra-varient;
 more concretely, show a diagram that doesn't commute)
eg of id -> power set for sets ($x \mapsto \set{x}$)
(have trivial EGs too)

the map $V \to V^{**}$ always exists, and is a natural trans.
of functors. (if you deal with topological vector spaces, life gets
more interesting) for algebraic vector spaces, it's always injective;
hence for finite-dimensional vector spaces, always isomorphism,
so it's a ``natural equivalence of functors''

$V = \Hom(K,V) \to \Hom(V^*,K^*) \to \Hom(V^*,K) = V^{**}$
this is good b/c it's purely categorical

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Duality of subsets

(this is a good example of a Galois connection)

(I think $S^* = \set{x\in V^* \st x(s)\leq 1 \forall s\in S}$ works)
(note that $S^{**}$ needn't equal $S$, but it always contains it,
 and $S^{***}=S^*$ always; this closure-type situation
 always occurs in Galois connections)
(this is useful in Economics, I think -- and not well-known in math)
(this makes rigorous the ``duality of polyhedra'', as in cube/octahedron
 etc. -- this latter makes a good example (try 2d first)
 -- but note that that's actually a different story!
 That is, ``put a vertex in the center of every face'' agrees with
 this duality, up to rescaling, for regular and semi-regular
 polytopes, but -not- for general ones!)

(proof of properties:
 dual is inclusion-reversing (pure logic);
 dual is convex (calculation);
 $S \subset S^{**}$ (pure logic);
 if S is convex, then $S^{**}=S$
   (trickiest bit -- several ways to show:
    concretely, if $x \nin S$ and $S$ is convex, then it's separated by
    -some- hyperplane, etc.
    more abstractly, convex means intersect of hyperplanes etc.)
 (okay, better: it's purely formal: double dual of half-space is the
 same half space, so they're closed. A convex set is an intersection
 of halfspaces, and intersection of closed is closed.)

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Free things

The similarity between sets and vector spaces is not accidental;
vector spaces are "free objects" in some sense (get the details right)
in particular, introduce adjoints, esp the Free/Forget adjunction

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What -are- duals?
(put in the epilogue)

involutions -- indeed, even an equality can be considered
a duality of a sort -- it's two different ways of looking at something.
(mention pontryagin duality)

