Page 17 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Theorem 2: Let f : A  B be a function. Then,
(a) for any subset S of B, f(f (S) )  S.
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(b) for any subset X of A, X  f (f(X)).
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Proof: We will prove (a) and you can prove (b). Let b " f(f (S) ). Then, by definition, 3 a  f (S)
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such that b = f(a). But a  f (S)  f(a)  S. That is, b  S.
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Thus, f(f (S) )  S.
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Now let us look at the most important way of producing new functions from given ones.
1.4.1 Composition of Functions

If f : A  B and g : C  D are functions and if the range of f is a subset of C, there is a natural way
of combining g and f to yield a new function h : A # D. Let us see how.
For each x  A, h(x) is defined by the formula h(x) = g(f(x)).
Note that f(x) is in the range of f, so that f(x)  C. Therefore, g(f(x)) is defined and is an element
of D. This function h is called the composition of g and f and is written as g o f. The domain of
g o f is A and its codomain is D. In most cases that we will be dealing with we will have B = C.
Let us look at some examples.

Example: Let f : R  R and g : R  R be defined by f(x) = x and g(x) = x + 1. What is
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g o f ? What is f o g ?
Solution: We observe that the range of f is a subset of R, the domain of g. Therefore, g o f is
defined. By definition, V x  R, g o f(x) = g(f(x)) = f(x) + 1 = x + 1.
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Now, let us find f o g. Again, it is easy to see that f o g is defined. V x  R,

f o g(x) = f(g(x)) = (g(x)) = (x + 1) .
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So f o g and g o f are both defined. But g o f  f o g . (For example, g o f(1)  f o g(l).)

Example: Let A = {1, 2, 3}, B = {p, q, r} and C = {x, y}. Let f : A  B be defined by f(1) = p,
f(2) = p, f(3) = r. Let g : B  C be defined by g(p) = x, g(q) = y, g(r) = y. Determine if f o g and g o
f can be defined.
Solution: For f o g to be defined, it is necessary that the range of g should be a subset of the
domain of f. In this case the range of g is C and the domain of f is A. As C is not a subset of A,
f o g cannot be defined.
Since the range of f, which is (p, r), is a subset of B, the domain of g, we see that g o f is defined.
Also g o f : A  C is such that
g o f(l) = g(f(1)) = g(p) = x,
g o f(2) = g(f(2)) = g(p) = x,
g o f(3) = g(f(3)) = g(r) = y.

In this example note that g is surjective, and so is g o f.
We now come to a theorem which shows us that the identity function behaves like the number
1  R does for multiplication. That is, if we take the composition of any function f with a suitable
identity function, we get the same function f.

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