f

by the rules

f

The assignment of hom(*,A) to any object A of

(fg)

Part c) was straightforward, although admittedly tedious.

The statement that Hom(W,-) is right exact is simply the statement that every homomorphism f from W to V'' necessarily lifts to a homomorphism from W to V if pi:V-->V'' is a surjective homomorphism. Such a lift could be found for vector spaces by taking a basis for W, sending each basis vector w to a preimage of f(w) under pi, and extending the resulting map by F-linearity.

The statement that Hom(-,W) is right exact is simply the statement that every homomorphism f from V' (contained in a vector space V) to W extends to a homomorphism from V to W. Such an extension is obtained by completing the basis for V' to a basis for V, and sending the additional basis vectors to arbitrarily chosen vectors in W.

In both cases, the existence of bases for vector spaces was used in a crucial way.

In the context of modules, one can find many counterexamples: for example, letting Z/nZ be the Z-module of integers modulo n, the functor Hom(Z/nZ,-) does not preserve exactness on the right, because (for example) the identiy map Z/nZ-->Z/nZ does not lift to a homomorphism Z/nZ --> Z. (There are no such homomorphisms, other than the trivial one!).

Similarly, Hom(-,Z) is not exact, as the homomorphism from nZ to Z sending n to 1 does not extend to a homomorphism from Z to Z, if n>1. (Such an extension would need to send 1 to 1/n.)

X = Z[x,y]/(x

the quotient of the polynomial ring in two variables with coefficients in Z, by the ideal generated by the relation. Indeed, to give a homomorphism from X to R is "the same thing" as specifying the images of x and y, which must satisfy (in R) the same algebraic relation as x and y satisfy in X. Of course, this phenomenon is completely general: given and set of algebraic equations in a finite number of variables, the functor which to a ring R associates the set of solutions to the system of equation with entries in R is a representable functor.

M = { (g,x) such that g x=x } (in G x X.)

Now count the cardinality of M in two different ways. On the one hand it is equal to the sum of FP(g), as G varies ove G; on the other hand it is equal to the sum of the cardinality of Stab(x), as x runs over X. Now evaluate the latter expression by using the counting formula for group actions.... Part b) could be reduced to part a), by consider the action of G on the set Y of ordered pairs of distinct elements of X. (I was surprised than many people, including those who were able to solve the harder part a, did not write a solution to part b.)

The proof that every group of order a power of p is solvable now follows by an induction on the size of the group, using the fact that G is solvable if and only if the quotient of G by its center is.

(p

(p

The group GL