Math 55a: A preview of ``abstract nonsense''
Suppose we have a sequence of vector spaces and linear maps
... >
V_{n1} >
V_{n} >
V_{n+1} >
...
The sequence is said to be ``exact at V_{n}''
if the kernel of the map
V_{n}>V_{n+1}
equals the image of the map
V_{n1}>V_{n}.
[As a consequence, the composite map
V_{n1}>V_{n+1}
must be the zero map.] The sequence is said to be ``exact''
if it is exact at each vector space with an incoming and outgoing arrow.
The simplest cases:

0 > V_{1} > 0
is exact iff V_{1} is the zero space.

0 > V_{1} > V_{2} > 0
is exact iff the map
V_{1}>V_{2}
is an isomorphism.

0 > V_{1} > V_{2}
> V_{3} > 0
is exact iff
V_{1}>V_{2}
is an injection,
V_{2}>V_{3}
is a surjection, and the image of the first map
equals the kernel of the second; that is, if and only if
the sequence is equivalent to
0 > U > V
> V/U > 0
for some vector space V with a subspace U
mapped into V using the restriction to U
of the identity map on V.
An exact sequence
0 > U > V
> V/U > 0
is called a ``short exact sequence''; an exact sequence
involving four or more vector spaces between the initial and final zero
is called a ``long exact sequence''. In any exact sequence
of finitedimensional vector spaces with an initial and final zero,
the dimensions of the even and oddnumbered vector spaces
in the sequence have the same sum; in other words,
the alternating sum of the dimensions
(a.k.a. the ``Euler characteristic'' of the sequence) vanishes.
[Check that this holds for the above cases of sequences of length
at most 3.]
(Much the same definitions are made for sequences
in other ``categories'', such as groups,
or modules over a given ring. For instance,
0 > Z > Z
> Z/NZ > 0
is a short exact sequence of commutative groups
if we use multiplication by N as the map
Z>Z.)
If the vector spaces V_{n}
in our exact sequence are finite dimensional
then the dual spaces V_{n}^{*}
form an exact sequence with the arrows going in the opposite direction:
... <
V_{n1} <
V_{n} <
V_{n+1} <
...
[A mathematician enamored with abstract nonsense
would express this fact by saying that ``duality
is an exact contravariant functor on the catogery
of finitedimensional vector spaces and linear maps'';
the canonical identification of the second dual
V^{**} of every finitedimensiona
vector space V with V would likewise
give rise to an ``exact covariant functor'' on the same category.]
For instance, if we dualize
0 > U > V
> V/U > 0
we get
0 < U^{*} < V^{*}
< (V/U)^{*} < 0
Which is to say that (V/U)^{*}
is the kernel of a surjective linear map from
V^{*} to U^{*}
obtained from the injection of U into V.
This map is none other than the restriction map from
linear functionals on V to linear functionals on U;
the kernel of this map consists exactly of those linear functionals
that vanish on all of U. So we recover the fact that
(V/U)^{*} is the annihilator
of U in V^{*},
and that the quotient of V^{*}
by its subspace (V/U)^{*}
is canonically identified with U^{*}.
How much of this still works if we drop the requirement that
V be finite dimensional?