Questions from students on 2020-01-23

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1. I don't really understand what a syzygy is and why its useful, or I
could do with some clarification on why we need free resolutions.

Answer: A "syzygy" is a relation between relations.  Suppose that M is
an R-module.  Let {F_i} be a free resolution of M.  That is, {F_i} is
an exact sequence

\ldots F_3 \to F_2 \to F_1 \to F_0 \to M \to 0

where all of the F_i are free.  So the generators of F_0 give
generators for M, the generators of F_1 give relations for M, the
generators of F_2 give relations among the relations of M (syzygies),
the generators of F_3 give relations among the the relations among the
relations (higher order syzygies), and so on.

We need free resolutions to define the Ext groups.  The group Ext^1 is
the only one we really need - it comes from cocycles modulo
coboundaries in F_2.  That is, Ext^1 is a quotient of a submodule of
the module of syzygies!

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2. Does homology appear in other areas of Maths like cohomology does?

Answer: I said in lecture is that cohomology is more prevalent - I
didn't mean to imply that homology is nonexistent!

For example, people in geometric combinatorics use degree theory to
work on the square peg problem.  Homology also comes up in topology
(of course), in differential geometry, and in related fields.

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3. In the proof of the UCT, you did not say why the s.e.s. splits.
May you say something about this?

Answer: This first requires that you understand the proof that the
homomorphism h is surjective.  Here is a sketch of both that and of
the existence of the splitting.

Let 

q \from Z_n \to H_n

be the quotient homomorphism.  Fix a splitting of the sequence

0 \to Z_n \to C_n \to B_{n-1} \to 0  (**)

This gives us a projection

p \from C_n \to Z_n.

Suppose now that \phi is a homomorphism from H_n(C) to Q. We define

\phi' = \phi \circ q \circ p

Check that

a) \phi' is a cocycle and

b) h([\phi']) = \phi

So define k \from Hom(H_n(C), Q) \to H^n(C; Q) by k(\phi) = [phi'].

Check that

c) k is R-linear.

Thus k is the desired section.  Note that the only choice we made was
of the splitting of the sequence (**).  However, varying this choice
can cause the section k to vary.

Exercise: Give an example justifying this last claim. 

4. Is co-(cohomology) the same as homology? (i.e. does taking the dual
of the dual give you back the same thing?).

Answer: This depends on the details of how you do things.  But suppose
we take the double dual of the give chain complex C.  With field
coefficients C** is naturally isomorphic to C, and thus their homology
groups are as well.  This does _not_ work for rings (commutative, with
unit) in general.

Exercise: Give an example justifying this last claim.