EQUIVALENCE RELATIONS AND FUNCTIONS.

1. Definition. A f : A → B is said to be compatible with an equivalence R on A if: (∀x, y ∈ A)(xRy → f(x) = f(y)). Given an R on A, we always have the function (called the “quotient ”) π : A → A/R defined by π(x) = [x], the equiva- lence of x (recall A/R is the 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 “”) 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 h([x]) of B, regard- less of which element of A we pick as representative of the [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 ∈ 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) = |x|

, induce the same equivalence relation on R, since:

f(x) = f(¯x) ↔ x2 =x ¯2 ↔ |x| = |x¯| ↔ 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 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 ↔ (∀x, y ∈ 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 ∈ Z.

Let B ⊂ R be the set: B = (−10, −1] ∪ {0} ∪ [1, 10). We can use the results above to construct a bijection h : R/ ∼→ B. For this it suffices to find a 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 ∈ 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 ∈ Z so that 10 ≤ x < 10 , or 1 ≤ 10 x < 10. In this case, set: f(x) = 10−nx ∈ B. (c) If x < 0, we have −10n+1 < x ≤ −10n for a unique −n −n n ∈ Z, or −10 < 10 x ≤ −1. In this case we set f(x) = 10 x ∈ 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: ∀x, y ∈ 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] = {0}) 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 \{0}:

v ∼ w ↔ (∃c > 0)(v = cw).

(Exercise: this is an equivalence relation.) We now want to define a bijection 2 2 h :(R \{0})/ ∼→ C, where C = {v ∈ R ; |v| = 1} is the unit circle, and p 2 2 |v| = 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 \{0}. Then h([v]) = f(v) defines 2 the sought-after bijection h :(R \{0})/ ∼→ C. It is natural to consider the function: v f : 2 \{0} → C, f(v) = . R |v|

(i) f is surjective. Note that (∀v ∈ C)(f(v) = v) (since |v| = 1 for v ∈ C). 2 Hence any vector v ∈ C is the of under f of a vector in R \{0} (namely v itself), so f is surjective.

Next we need to show:

2 (∀v, w ∈ R \{0})(v ∼ w ↔ f(v) = f(w)).

2 Proof. Let v, w ∈ R \{0} be arbitrary. First, assume f(v) = f(w), or v w |v| = |w| . Then letting c = |v|/|w| 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 |v| = c|w| and dividing these two equalities we see that: |v| = |w| , 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 lines defines the same 2 unoriented direction.) The equivalence relation in R \{0} is:

v ∼ w ↔ (∃c 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+ = {v ∈ R \{0}|v = (x1, x2), x2 ≥ 0}\{(−1, 0)}.

2 Find a function f : R \{0} → C+ with the properties: f(v) = v for v ∈ C+ (so f is onto) (ii) f defines the same equivalence relation as ∼: v ∼ w ↔ f(v) = f(w).

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