Math 55a: A preview of ``abstract nonsense''

Suppose we have a sequence of vector spaces and linear maps ... --> V_n-1 --> 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_n-1-->V_n. [As a consequence, the composite map V_n-1-->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 --> V₁ --> 0 is exact iff V₁ is the zero space.
0 --> V₁ --> V₂ --> 0 is exact iff the map V₁-->V₂ is an isomorphism.
0 --> V₁ --> V₂ --> V₃ --> 0 is exact iff V₁-->V₂ is an injection, V₂-->V₃ 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 finite-dimensional vector spaces with an initial and final zero, the dimensions of the even- and odd-numbered 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 some 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_n-1 <-- 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 finite-dimensional vector spaces and linear maps''; the canonical identification of the second dual V^** of every finite-dimensiona 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?