Study Guide for Midterm Exam

The midterm exam problems will be very similar to some of the problems listed below. The exam will consist of approximately five such problems.

Module Theory

Problem 1

Let R be a commutative ring (with unity). We have shown that for every R-module M there is an R-module isomorphism HomR(R,M)M. Now show more generally that for each fixed positive integer n there is also an R-module isomorphism
HomR(i=1nR,M)i=1nM.


Problem 2

Let R be a commutative ring (with unity), X a finite set, and F(X) the free R-module on X. Prove there is an R-module isomorphism HomR(F(X),R)F(X).


Problem 3

Suppose f:NP is a (S,T)-bimodule morphism.

  1. Show that for every (R,S)-bimodule M there is an (R,T)-bimodule morphism
    1Mf:MSNMSP,
    defined on simple tensors by mnmf(n).

  2. Prove that if f is surjective, then so is 1Mf.

  3. Show there are isomorphisms of abelian groups Z2ZZZ2 and Z2ZQ0, and then explain why this proves tensor product does not in general preserve injections.


Problem 4

Let M1 and M2 be R-modules and A1M1 and A2M2 be submodules. Prove that A1×A2 is (isomorphic to) a submodule of M1×M2 and that there is an R-module isomorphism
(M1×M2)/(A1×A2)(M1/A1)×(M2/A2).


Problem 5

Let V be a finite-dimensional vector space over a field F and let v,wV be nonzero vectors. Prove that vw=wv in VFV if and only if w=av for some aF.

Category Theory Problems

Problem 6

Let F,G:CSet be functors and τ:FG be a natural transformation between those functors. We say τ is:

  • a natural bijection if for every cC the set map τc:F(c)G(c) is a bijection;
  • a natural isomorphism if there is a natural transformation η:GF such that ητ=1F and τη=1G.

Prove that τ is a natural bijection if and only if it is a natural isomorphism.


Problem 7

Suppose C is a category that has pullbacks and a terminal object, t.

  1. Prove that C has all (binary) products.
  2. Prove that C has all equalizers.

Problem 8

Suppose C is a category that has all equalizers.

  1. The map that assigns to each pair of parallel arrows f,g:ab the equalizer object Eq(f,g) is the object function of a functor to C. What is the arrow function of that functor?
  2. The equalizer functor above is right adjoint to a certain diagonal (or constant) functor from C. Describe:
    • a) the other functor;
    • b) the natural bijection of the adjunction; and
    • c) the unit and counit of the adjunction.

Problem 9

To each category C we can associate an opposite category Cop. The objects of Cop are the objects of C, and the arrows of Cop are arrows fop in one-to-one correspondence ffop with the arrows f of C. For each arrow f:ab in C, the corresponding arrow is fop:ba in Cop.

  1. Suppose F:CD is a functor. Show that the maps cF(c) and fop(F(f))op together define a functor Fop:CopDop.

  2. Consider a functor F:CopD. If we write F(f) for F(fop), show how the functor F can be expressed directly in terms of the original category C as maps (both denoted F) on objects and arrows in C. How does F interact with composition of arrows?

    Such maps are sometimes called contravariant functors from C to D, in which case our usual functors are called covariant functors.

  3. For each fixed object b in a category C, show how to define a functor

    HomC(,b):CopSet.

    This functor is sometimes called the contravariant hom-functor (associated to the object b).


Problem 10

Suppose C is a category, F:CSet is a functor and rC is an object of C, and let Hr=HomC(r,) be the hom-out functor associated to the object r. Each natural isomorphism α:HrF is called a representation of the functor F and a universal property for the object r.

Suppose α:HrF and β:HsF are two natural isomorphisms.

  1. Show there is a unique arrow f:rs in C such that β=αH(f).[1]
  2. Show that the unique arrow f in part (a) is an isomorphism; i.e., there is an arrow g:sr in C such that gf=1r and fg=1s.

Because of this result, we say that "representations of functors are unique up to unique isomorphism," and also "objects that satisfy a universal property are unique up to unique isomorphism."


  1. This is if we're considering H as a contravariant functor from C to SetC. If we're considering H as a functor from Cop to Set, then this should be H(fop). ↩︎