An Important Example of an Equivalence Relation and Well-Definedness
Suppose X and Y are sets and that f : X → Y is a function. Let us define a relation ∼ on X by
x1 ∼ x2
if and only if
f (x1 ) = f (x2 ).
(1)
Review Exercise 1. Show that ∼ is an equivalence relation.
Proof. The definition of equivalence relation says that we must prove that ∼ is reflexive, symmetric and
transitive.
(i) Reflexive: Since f (x) = f (x) for every x ∈ X, the definition (1) implies that x ∼ x.
(ii) Symmetric: If x1 ∼ x2 , then f (x1 ) = f (x2 ). But then f (x2 ) = f (x1 ), and hence x2 ∼ x1 , as desired.
(iii) Transitive: If x1 ∼ x2 and x2 ∼ x3 , then f (x1 ) = f (x2 ) and f (x2 ) = f (x3 ). But then f (x1 ) = f (x3 ),
and hence x1 ∼ x3 , as desired.
Remark. Notice that in the proof above we used the fact that = is an equivalence relation, and hence is
itself reflexive, symmetric and transitive.
Review Exercise 2. If the function f that we started with is one to one, what can we say about ∼ ?
Since ∼ is an equivalence relation, we define equivalence classes and the set of all equivalence classes for ∼ .
Definition. (i) [x] = {x̂ ∈ X : x̂ ∼ x} is called the equivalence class of ∼ represented by x, or, less
formally, the equivalence class of x.
(ii) X̂ = {[x] : x ∈ X} is the set of all equivalence classes of ∼ .
Remarks. (i) Note that a given equivalence class can have many representatives. In fact, [x1 ] = [x2 ] if
and only if x1 ∼ x2 . (Show!)
(ii) Precisely speaking, equivalence classes are subsets of X, so X̂ is a set of subsets of X. Usually, though,
we just think of X̂ as a new set of elements and don’t stop to think of those elements as sets.
Suppose we set
fˆ([x]) = f (x)
for
[x] ∈ X̂.
(2)
Review Exercise 3. Show that fˆ: X̂ → Y is a well-defined, one to one function.
Proof. We have two things to do here. First, we must show that fˆ is well-defined, i.e., that it makes sense.
Isn’t that obvious? No! Looking carefully, we see that the definition of fˆ applied to [x] depends on x,
the representative we chose for the equivalence class [x]. As remarked above, [x] can have more than one
representative, i.e., we can have [x] = [x0 ] for x 6= x0 . If that’s true, can we be sure that fˆ([x]) is defined the
same way as fˆ([x0 ])? Well, if [x] = [x0 ], then x ∼ x0 , as noted above. But the definition of ∼ then yields
that f (x) = f (x0 ). But now using (2), we have
fˆ([x]) = f (x) = f (x0 ) = fˆ([x0 ]),
exactly what we need.
What remains is to show that fˆ is one to one. Recall that this means that if fˆ([x1 ]) = fˆ([x2 ]), then
[x1 ] = [x2 ]. But again using (2), we have that
f (x1 ) = fˆ([x1 ]) = fˆ([x2 ]) = f (x2 ),
and hence, using (1), that x1 ∼ x2 , and so [x1 ] = [x2 ], as desired.
Remark. This example is very important in algebra and we will see it in action when we study the First
Isomorphism Theorem a little later on.
c R. Kubelka 2013