EQUIVALENCE RELATIONS AND FUNCTIONS. 1. Definition. A function f : A ! B is said to be compatible with an equivalence relation R on A if: (8x; y 2 A)(xRy ! f(x) = f(y)): Given an equivalence relation R on A, we always have the function (called the \quotient projection") π : A ! A=R defined by π(x) = [x], the equiva- lence class of x (recall A=R is the set of all equivalence classes, so A=R is a \family of sets".) Problem 1. If f : A ! B is compatible with an equivalence relation R on A, show that the function (or \map") h : A=R ! B given by h([x]) = f(x) is well-defined, and satisfies h ◦ π = f. (\Well-defined” here means we get the same element h([x]) of B, regard- less of which element of A we pick as representative of the equivalence class [x]). Going in the other direction: any function f : A ! B defines an equiva- lence relation Rf in its domain A by the rule: xRf y $ f(x) = f(y); for x; y 2 A: Problem 2. Prove that this is an equivalence relation, and that f is compat- ible with Rf . Example 1. Two different functions f; g with the same domain A may induce the same equivalence relation Rf = Rg on A. For example, 2 f : R ! R; f(x) = x and g : R ! R; g(x) = jxj , induce the same equivalence relation on R, since: f(x) = f(¯x) $ x2 =x ¯2 $ jxj = jx¯j $ g(x) = g(¯x): Problem 3. Show that the functions sin x and cos x (both from R to R) define the same equivalence relation on R; describe a typical equivalence class (say, [π=3].) It is a common situation in Mathematics that one needs to idenfity the abstractly defined set A=R in terms of a more \concrete" set B, ideally by exhibiting a bijection between them. If we can find a set B and a function f : A ! B in such a way that R = Rf , from the preceding we'll have a function h : A=R ! B; so we need to find conditions for the injectivity and surjectivity of h. 1 Problem 4. Let f : A ! B be compatible with an equivalence relation R on A. Let h : A=R ! B be defined as above: h([x]) = f(x). Prove that: h is injective $ (8x; y 2 A)(f(x) = f(y) ! xRy): In particular, this shows that for any function f : A ! B, the function h : A=Rf ! B (defined by the equivalence relation Rf ) is always injective. Problem 5. Prove that if f : A ! B is surjective, then h : A=Rf ! B is also surjective. (From Problem 4 we already know it is injective, hence it is always bijective.) Conclusion: if R is an equivalence relation on A and we can find a set B and a function f : A ! B so that R = Rf (that is, f induces the same partition of A as R does) and f is surjective, then the function h : A=R ! B given by h([x]) = f(x) is well-defined, and a bijection. Example 2. Consider the equivalence relation ∼ on R defined by: n x ∼ y $ x = 10 y; for some n 2 Z: Let B ⊂ R be the set: B = (−10; −1] [ f0g [ [1; 10): We can use the results above to construct a bijection h : R= ∼→ B. For this it suffices to find a surjective function f : R ! B so that Rf and ∼ are the same equivalence relation on R. (The latter means: f(x) = f(y) $ x ∼ y, for all x; y 2 R.) Let f : R ! B be defined as follows; (a) f(0) = 0; (b) If x > 0, there is a n n+1 −n unique n 2 Z so that 10 ≤ x < 10 , or 1 ≤ 10 x < 10. In this case, set: f(x) = 10−nx 2 B. (c) If x < 0, we have −10n+1 < x ≤ −10n for a unique −n −n n 2 Z, or −10 < 10 x ≤ −1. In this case we set f(x) = 10 x 2 B. Problem 6. Show that f : R ! B is surjective. Problem 7. Show that Rf and ∼ are the same equivalence relation on R, that is: 8x; y 2 R x ∼ y $ f(x) = f(y): Hint: consider separately the cases (i)x = y = 0; (ii) x > 0; y > 0; (iii) x < 0; y < 0. This is possible since all the real numbers in a given equivalence class (other than [0] = f0g) are positive, or all are negative, and f(x) has the same sign as x, for any x (including f(0) = 0.) Thus the corresponding function h : R= ∼→ B given by h([x]) = f(x) is well-defined, and a bijection. 2 Example. It is geometrically clear that \oriented directions in the plane are parametrized by the unit circle", but it is useful to make this more 2 precise. Two nonzero vectors in R define the same oriented direction if they one can be obtained from the other by multiplying by a positive scalar. 2 This suggests setting up the equivalence relation on R n f0g: v ∼ w $ (9c > 0)(v = cw): (Exercise: this is an equivalence relation.) We now want to define a bijection 2 2 h :(R n f0g)= ∼→ C, where C = fv 2 R ; jvj = 1g is the unit circle, and p 2 2 jvj = x1 + x2 if v = (x1; x2). From the discussion above, it is enough to 2 find a function f : R ! C such that (i) f is surjective and (ii) f and ∼ 2 define the same equivalence relation on R n f0g. Then h([v]) = f(v) defines 2 the sought-after bijection h :(R n f0g)= ∼→ C. It is natural to consider the function: v f : 2 n f0g ! C; f(v) = : R jvj (i) f is surjective. Note that (8v 2 C)(f(v) = v) (since jvj = 1 for v 2 C). 2 Hence any vector v 2 C is the image of under f of a vector in R n f0g (namely v itself), so f is surjective. Next we need to show: 2 (8v; w 2 R n f0g)(v ∼ w $ f(v) = f(w)): 2 Proof. Let v; w 2 R n f0g be arbitrary. First, assume f(v) = f(w), or v w jvj = jwj . Then letting c = jvj=jwj we have v = cw with c > 0, so v ∼ w. For the converse, assume v ∼ w. Then v = cw for some c > 0, so v w jvj = cjwj and dividing these two equalities we see that: jvj = jwj , or f(v) = f(w): This ends the proof. Exercise. Now try to repeat the above to parametrize the set of unori- ented directions in the plane (any family of parallel lines defines the same 2 unoriented direction.) The equivalence relation in R n f0g is: v ∼ w $ (9c 6= 0)(v = cw): (We no longer require c to be positive.) Show that this is an equivalence rela- tion. It is geometrically natural to try to parametrize the set of equivalence 3 classes (which is the set of unoriented directions| by the upper semicircle, including its right endpoint: 2 C+ = fv 2 R n f0gjv = (x1; x2); x2 ≥ 0g n f(−1; 0)g: 2 Find a function f : R n f0g ! C+ with the properties: f(v) = v for v 2 C+ (so f is onto) (ii) f defines the same equivalence relation as ∼: v ∼ w $ f(v) = f(w). 4.