1 Equivalence Relations

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1 Mitchell Faulk June 12, 2014 Equivalence Relations and Quotient Spaces 1 Equivalence Relations Recall that a relation on a set X is a subset S of the cartesian product X X. Problem Set 2 has several examples of relations, so I will not spend time on examples here. Instead, I would like to discuss special types of relations, called equivalence relations, and how to deal with these special relations in mathematical constructions. As an example, I will introduce the notion of a quotient vector space. Let me start by discussing an equivalent formulation of a relation on a set X, namely, as a map X X {0, 1}. We can think of a relation S on X as defining a map where R S : X X {0, 1} R S (x, y) = { 1, (x, y) S 0, (x, y) / S In the case that R S (x, y) = 1, we say that x is related to y and we write this statement as xr S y. In the other case, the elements x and y are unrelated. The previous paragraph implies that we have a map R : {relations on X} Set(X X, {0, 1}) S R S where Set(X X, {0, 1}) is the collection of maps X X {0, 1}. One can show that this map R is a bijection, and thus we have an alternate, yet equivalent way of thinking about a relation on a set X, namely as a map from X X {0, 1}. Exercise 1.1. Show that the map R is a bijection by giving an inverse. Remark 1.2. More generally, for any set D, we may construct a bijection {subsets of D} Set(D, {0, 1}). The set of subsets of D is usually denoted P(D), and is called the power set of D. If D has finite cardinality D, the previous observation implies that the cardinality of P(D) is P(D) = 2 D. This is somewhat of a common technique in mathematics: to view one construction as another equivalent construction. Having several different viewpoints can be somewhat of an advantage, as each different viewpoint allows for 1

2 a different approach to a proof. On the other hand, having too many different viewpoints can make certain things confusing, since it might not be clear which viewpoint is best in which situation. Moreover, constantly switching viewpoints can be confusing in itself. Nevertheless, I tend to believe in the slogan the more viewpoints, the merrier. In the spirit of the slogan, we now give an equivalent formulation for an equivalence relation on a set X. In the end, we will see that giving an equivalence relation on X is the same as specifying a partition of the set X. Recall that a relation is called an equivalence relation if it is reflexive, symmetric, and transitive. (These three terms are defined in Problem Set 2.) If is an equivalence relation on a set X, then for any element x X, we may introduce the equivalence class of x, denoted [x] and defined by [x] := {y X : x y}. Note that [x] is a subset of the set X, namely, the subset consisting of all elements of X that are related to x under the equivalence relation. We then write X/ for the collection of all equivalence classes, that is, X/ := {[x] : x X}. We have a natural surjective map π : X (X/ ) which sends an element x to its equivalence class [x], that is, π(x) = [x]. Example 1.3. For a positive integer n, define a relation on Z by the rule x y if there is an integer k such that x = y + kn. (Morally, one should think of this relation as saying that x is related to y if x and y have the same remainder mod n.) It follows directly from the definition that the equivalence class of x Z is given by [x] = {x + kn : k Z} = {..., x 2n, x n, x, x + n, x + 2n,...}. We let Z n denote the set of equivalence classes under this relation. It is a rewarding exercise to show that the cardinality of the set Z n is finite and equal to n. Exercise 1.4. Show that the cardinality of Z n is n. The previous example hints at a phenomenon that occurs in general: the set of equivalence classes form a partition of the parent set. (The term partition is defined in Problem Set 2.) More precisely, we have the following proposition, whose proof we leave as an exercise. Proposition 1.5. For an equivalence relation on a set X, the set of equivalence classes form a partition of the set X. 2

3 Reformulated in another way, Proposition 1.5 says that we have a map {equivalence relations on X} {partitions of X} which sends an equivalence relation to the partition given by its equivalence classes. One could hope that this map is a bijection, which would mean that giving an equivalence relation on a set X is the same as giving a partition of a set X. And in fact, this is the case, as the following theorem asserts, which I let you prove in Problem Set 2. Theorem 1.6. Let X be a set. Then there are mutually inverse maps {equivalence relations on X} {partitions of X}. 2 Well-defined maps Let us motivate this section with a question. Suppose that we have a set X with an equivalence relation. Suppose also that we have a map f : X Y from X into another set Y. We can ask: When does the map f give rise to a map f : (X/ ) Y from the set of equivalence classes of X into Y? More precisely, we are interested in when we can define f by the rule f([x]) = f(x). To see why this is even a question at all, let us consider an example. Let f : Z Z be the identity map. We can ask: Does f induce a map f : Z 2 Z? Because we are using such a nice map, namely the identity map, one would guess that the answer to this question is yes. However, this is certainly not the case! Indeed, note that f maps 1 1 and 3 3. However, in Z 2, the elements 1 and 3 represent the same equivalence class, meaning that [1] = [3] in Z 2. Thus, if f were a function, we would have that 1 = f(1) = f([1]) = f([3]) = f(3) = 3, which is clearly false. So f does not induce a map f in this case. If we examine what went wrong in the previous example, we are led naturally to introduce the following notion. Definition 2.1. Let X be a set with an equivalence relation and let f : X Y be a function from X into another set Y. We say that f induces a well-defined map f : (X/ ) Y if and only if the following property is satisfied by the map f: whenever x and x are elements of X such that x x, we have f(x) = f(x ). If this property is satisfied, then the map f is defined by the rule f([x]) = f(x) for x X. 3

4 Exercise 2.2. Let Z n denote the integers mod n. Let + : Z Z Z denote the addition map on Z. For an integer y Z, the mapping x x + y defines a map R y : Z Z. Show that the map f := π R y : Z Z n induces a well-defined map f : Z n Z n. This shows that the binary operation given by : Z n Z n Z n ([x], [y]) [x] [y] := [x + y] is well-defined in the first slot. To show that it is well-defined in the second slot, you should show that the map g := π L y : Z Z n induces a well-defined map g : Z n Z n, where L y : Z Z is the map defined by L y (x) = y + x. The upshot of this exercise is that we can equip Z n with a binary operation : Z n Z n Z n induced by addition on Z. As a consequence of this, it is easy to see that (Z n,, [0]) is a group. 3 Quotient vector spaces Let V be a vector space over the field k and let U be a subspace of V. From this data, we will construct a new vector space V/U called the quotient space whose vectors are equivalence classes of vectors from V and whose operations of addition and scalar multiplication are induced by the corresponding operations on V. The idea of the quotient space construction is to collapse the subspace U to a single vector. That is, we should view all elements of U as representing the same vector, that is, as equivalent vectors under some appropriately defined equivalence relation. More precisely, define the following relation on the set V : We say that v w if there is a vector u U such that v = w + u. We leave it as an exercise to show that indeed defines an equivalence relation. Exercise 3.1. Show that is an equivalence relation. We let V/U denote the set of equivalence classes of vectors under this equivalence relation, that is, V/U = {[v] : v V }. This means that the set V/U has as its elements subset of V. Our goal is to endow this collection of subsets with the structure of a vector space. First, we might ask ourselves, what is the zero vector in this vector space? The following exercise hints at the answer to this question. Exercise 3.2. Show that u 0 for any u U. 4

5 Exercise 3.2 shows that any vector u U is equivalent to the zero vector 0 V. This implies that for u U, the equivalence class [u] which is the subspace U itself will be the zero vector in our new space. We might next ask ourselves: What are the operations of addition and scalar multiplication on V/U? Surely these operations should somehow be induced by the corresponding operations from V. Indeed, this is the case, and it is natural to define addition and scalar multiplication in the following manner. We leave it to the reader to verify that the following operations are in fact well-defined. We define vector addition by and scalar multiplication by V/U V/U V/U ([v], [w]) [v + w] k V/U V/U (λ, [v]) [λv]. Exercise 3.3. Show that the maps of addition and scalar multiplication on V/U are well-defined. The set V/U together with the above operations of addition and scalar multiplication is called the quotient of V by U. One often refers to this space simply as the quotient space or as V mod U. Exercise 3.4. Let V be a finite-dimensional vector space over k and let U be a subspace of V. Let π : V V/U denote the natural projection map. (a) Show that π : V V/U is a linear transformation. (b) Show that ker π = U. (c) Show that dim(v/u) = dim(v ) dim(u). Exercise 3.5. Let V and U be vector spaces over k (not necessarily finitedimensional). Let T : V U be a surjective linear transformation. Let W = ker T. Let T : V/W U be the map defined by T ([v]) = T v. (a) Show that T : V/W U is a well-defined map. (b) Show that T : V/W U is a linear transformation. (c) Show that T : V/W U is an isomorphism of vector spaces (that is, show that it is both injective and surjective). (This is called the first isomorphism theorem.) 5

6 Exercise 3.6. Let V be a vector space over k and let T : V V be a linear operator. For a subspace W of V, let T (W ) = {T w : w W }. Let U be a subspace of V. Show that the map induces a well-defined map if and only if T (U) U. π T : V V/U π T : V/U V/U 6