42 NOTESExample 7.2. Suppose that G is any group. Then there is a categorywhich we called G with one object ∗ and one morphism g : ∗ → ∗ forevery element g ∈ G. The composition of morphisms is defined to begroup multiplication:∗h−→ ∗g−→ ∗ gives g ◦ h = gh by definition.Then composition is associative:(f ◦ g) ◦ h = (fg)h = f(gh) = f ◦ (g ◦ h)The identity arrow is given by the identity of the group:id∗= eand we checked that it is the identity:id∗◦ f = e ◦ f = ef = f = f ◦ id∗Then we did an example of the example: Suppose G = &Z/2, +'.Then we get a category C+with one object ∗ and two morphisms0, 1 : ∗ → ∗ with composition law given by the chart:+0 10 0 11 1 0Since this is an specific example of the previous example, it is a cate-gory. The identity of ∗ is id∗= 0.Finally, we started the example of the multiplication rule for compo-sition: The category C×has one object ∗, two morphisms 0, 1 : ∗ → ∗with composition given by the chart:× 0 10 0 01 0 1The identity of ∗ in this category is id∗= 1.Question 7.3. Describe all categories with one object ∗ and two mor-phisms a, b : ∗ → ∗. In particular, are they all isomorphic to the abovetwo examples?MATH 47A FALL 2008 INTRO TO MATH RESEARCH 437.1.1. examples. On Monday we did this example. Suppose you have acategory with two objects: A, B and four arrows: A → A, A → B, B →A, B → B. The question is: What are these arrows? What should theybe called? What are their compositions?The answer is(1) Each object has an identity. So the two loops must be:idA: A → A, idB: B → B.(2) Call the other two arrows f : A → B, g : B → A.(3) The compositions must be the identities:f ◦ g = idB, g ◦ f = idAthe reason is that they cannot be anything else:g ◦ f : Af−→ Bg−→ ASince g ◦ f is an arrow from A to A it must be idA.The other example I started but did not finish was:Suppose you have a category with two objects: A, B and five arrows:A → A, A → B, B → A and two loops at B: B → B. The question is:What are these arrows? What should they be called? What are theircompositions?(1) We have the identities idA, idB(2) We have two arrows f : A → B, g : B → A.(3) We have one more arrow h : B → B.(4) The composition g ◦ f : A → A must be the identity of A sinceit can’t be anything else.(5) The composition f ◦ g : B → B is either idBor f ◦ g = h.(6) This implies that h ◦ h = h:h ◦ h = (f ◦ g) ◦ (f ◦ g) = f ◦ idA◦ g = f ◦ g = h.(7) All other compositions are uniquely determined.44 NOTES7.2. posets. A poset is a set P with transitive, anti-reflexive relation<. In other words, a < b < c implies a < c and a < b implies b ! a.We have to take the ≤ relation to get a category.7.2.1. integers with usual ordering. I started with an infinite categoryZ. This has objects all integers. I.e., each integer is one object. Thereis a unique morphism a → b if a ≤ b. If a > b there is no arrow a → b:(1) Ob(Z) = Z, i.e., the objects are · · · − 2, −1, 0, 1, 2, · · ·(2) Ar(Z): a → b if and only if a ≤ b.To get a finite category, take only a finite set of integers. For ex-ample take just 1,2,3. This makes a category with three objects and 6morphisms:• •1 f 2•g 3g ◦ fThis drawing has redundant information which I want to suppress.First, each object has an identity loop. So, we don’t draw them.We know they are always there and we know what their names are(id1, id2, id3). Also, the composition of the two basic arrows f, g isnecessarily present by definition of a category. We also know its name(g ◦ f). So, we don’t draw it. What I draw is:• −→ • −→ •Some people like to draw objects as points. Other people like to writethe names of the points:1 −→ 2 −→ 37.2.2. Hasse diagram. A Hasse diagram is a graph drawn in a planewhere all the edges are going up-down. There are no horizontal edges.When there is an edge connecting two points a, b with a below and babove this means a < b. By transitivity a < b < c gives a < c. In otherwords, x < y if there is a path from x to y which is always going