FINITE CATEGORIES WITH TWO OBJECTSTHE AUTHOR: MEAbstract. In this paper we will examine the possible configurations for finitecategories with two objects. In particular, we will look at the case when thereis exactly on morphism from B to A.A paper should have:(1) Title(2) Author(3) Abstract: a very short description of the paper. It can be vague.(4) Introduction: a more detailed description of the content.(5) The actual paper.(6) References.Optional are: Date, table of contents, information about the author.IntroductionSuppose that C is a category with two objects A, B. Then we want to examinethe three separate structures: the endomorphism sets of A and B and the homset Hom(A, B). Endomorphisms are morphisms from an object to itself. We willexamine two kinds of endomorphism sets: those which are groups and those whichhave annihilators. Then we will look at the consequences for their actions onHom(A, B). Actually, I won’t do that. This is just an example of what you mightwrite which sounds nice. Intoductions should have some overblown hype withpromises of interesting reading. Introductions should also have an outline of thecontents.In the first section we will review some basic definitions. In the second sectionwe examine the structure of the endomorphism sets End(A) = Hom(A, A). In thethird and last section we see how this affects the structure of the set Hom(A, B).1. DefinitionsRecall that a category consists of “objects” and “morphisms” (also called “ar-rows” or “homomorphisms”) with a composition law which is associative with units.More precisely we have the following. (For more details, see [ML98].)Definition 1. A category C has a collection Ob(C) of objects, A, B, C, etc and forany two objects A, B a set Hom(A, B) of arrows f : A → B so that(1) Composition of arrows f : A → B, g : B → C is defined giving an arrowg ◦ f : A → C(2) Composition is associative: (f ◦ g) ◦ h = f ◦ (g ◦ h)(3) Each object A has an identity morphism idA: A → A so that id ◦ f =f ◦ id = f.12 THE AUTHOR: MEAt this point you could put in an example. You need to put in a short definitionand example so that people know what you are talking about. Even if they alreadyknow.Here is a very simple example:Example 2. Take 3 objects: A, B, C and put in exactly 6 arrows: A → A, B →B, C → C, A → B, B → C, A → C. Composition is uniquely determined. So,associativity is automatic. The identity morphisms are the unique arrows fromeach object to itself.The first question is: What can you say if there are two morphisms A → A?You can ask rhetorical questions. Then you might need new definitions to make iteasier to talk about. Also this gives you a feeling that you are “creating math” or“designing” something. To explore this question we use the definition: If A is anobject in a category C, we define an endomorphism of A to be an arrow A → A.Note that the set of endomorphisms of A is closed under composition and there isan identity: idA. We call the set of endomorphisms of A, End(A).2. EndomorphismsNote that I haven’t actually answered the question. I only made a bunch ofdefinitions and new words.Suppose that End(A) has two elements. Then one of them is the identity idAand the other is some other arrow f : A → A. Three of the possible compositionsare determined:f ◦ idA= f, idA◦ f = f, idA◦ idA= idAThis leaves only one composition which is not determined: f ◦ f. There are twochoices:f ◦ f = f, f ◦ f = idA.Theorem 3. Both of these are possible.Proof. To show that something is possible we just need one example. We needsomething which we know is a category. The first example is the category of setsand bijections. Since the composition of bijections is a bijection and the identitymapping of a set to itself is a bijection, this is a category. Let A be the set withtwo elements A = {a, b}. Then A has two endomorphisms in this category: idAand f which switches the two elements:f(a) = b, f(b) = a.In this case f ◦ f = idA.Another example is a category in which the functions are labeled with numbersand composition is given by multiplication. Then each object needs to have anendomorphism labeled “1” and the set of numbers needs to be closed under multi-plication. In this example, we can define End(A) = {0, 1} Since this set of numbersis closed under multiplication and includes the number 1, we get a category withone object and two morphisms. Composition of the nonidentity arrow with itself is0 · 0 = 0So, f ◦ f = f is possible. So, both choices for f ◦ f are possible. FINITE CATEGORIES WITH TWO OBJECTS 3Next, suppose that are three endomorphisms of A. Since one of them must bethe identity, we have:End(A) = {idA, f, g}.There are four undetermined compositions:f ◦ f, g ◦ g, f ◦ g, g ◦ fWe could write down all the possibilities. But this would be very tedious and notvery instructive. Instead, I will get a more interesting conceptual partial answer.We divide into cases depending on whether multiplication (composition) satisfiescancellation or not.We say that multiplication by f satisfies left cancellation iff ◦ x = f ◦ y ⇒ x = y(in other words, you can cancel f when it is on the left). In this case,f ◦ idA, f ◦ f, f ◦ gmust all be different. In particular, one of these compositions must be equal to idA.So,Lemma 4. If End(A) is finite and f ∈ End(A) satisfies left cancellation then fhas a right inverse (a morphism h : A → A so that f ◦ h = idA.If both f and g satisfies left cancellation then all elements will have right inversesmaking End(A) into a group. But these is only one group of order 3, namely thecyclic group. So, we have the following.Theorem 5. If f and g both satisfy left cancellation then they are inverse to eachother: f ◦ g = g ◦ f = idAand f ◦ f = g, g ◦ g = f.In the other extreme, we could have an “annihilator” also called 0.Definition 6. An annihilator is defined to be any endomorphism 0 : A → A sothat 0 ◦ h = 0 = h ◦ 0 for all endomorphisms h of A.If A has an annihilator, say g = 0, then the remaining endomorphism f has oneundetermined composition: f ◦ f. This could be either 0, idAor f.Going back to Theorem 3, we note that the two possible structures of the casewhen End(A) has two elements are(1) End(A) is a group with two elements. Thus it is the cyclic group (Z/2, +)(2) End(A) has an identity and an annihilator. In this case End(A) is themultiplicative structure: (Z/2, ×).We note