MATH 301
LECTURE ON FUNCTIONS
DAVID MEREDITH
(1) (Simple definition) Suppose A and B are sets. A function f from A to B,
f
(2)
(3)
(4)
(5)
denoted either f : A → B or A −
→ B assigns a value f (a) ∈ B to each value
a ∈ A.
(Fancy definition–needed since “assign” is a little vague) Let A and B be
sets. A relation on A × B is a subset R ⊆ A × B. Say a is related to
b if (a, b) ∈ R. A graph is a relation G ⊆ A × B containing exactly one
pair (a, b) ∈ G for each a ∈ A. The function associated with the graph is
f (a) = b if and only if (a, bj) ∈ G. By defniition f (a) has a unique value
for each a ∈ A.
Which definition makes more sense to you. Which would be easier to use
in a proof?
A function f : A → B has three parts: the domain A, the codomain B
and f , for which there is no common name. We will call f the rule. All
functions have three parts, and two functions are the same only if all three
parts are the same. The function f : N → Z, f (n) = 2n is not the same
function as g : Z → Z, g(z) = 2z, because the domains are different.In
calculus (and analysis) functions are often identified with their rules, and
two functions with the same rule are considered the same. But in algebra
and topology it is important to specify the domain and codomain as well
as the rule when defining a function.
Examples:
f : Z → Z f (x) = x2
g : N → Z g(x) = x2
h : N → N h(x) = x2
k : R → R k(x) = x2
m : R≥0 → R m(x) = x2
n : R≥0 → R≥0 n(x) = x2
(6) To understand these notes, think of examples using sets with one or two
elements. That’s all you need to illustrate all of the ideas.
(7) Extended notation: if f : A → B is a function and C ⊆ A, then f (C) =
{f (c) : c ∈ C} ⊆ B.
(a) The image of f is im(f ) = f (A)
(b) In the examples above:
(i) im(f ) = the square integers including 0.
Date: November 3, 2011.
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DAVID MEREDITH
(ii)
(iii)
(iv)
(v)
(vi)
im(g) the non-zero square integers.
im(h) = the non-zero square integers.
im(k) = R≥0
im(m) = R≥0
im(n) = R≥0
.
(8) A function f : A → B is 1-1 or injective or monomorphic if f (a) = f (a0 )
implies a = a0 .
(∀a ∈ A)(∀a0 ∈ A)(f (a) = f (a0 ) =⇒ a = a0 )
The functions g, h, m and n are injective.
(9) A function f : A → B is onto or surjective or epimorphic if f (A) = B. To
prove that f is onto, show for each b ∈ B you can find a ∈ A such that
f (a) = b.
The function n is surjective.
f
g
(10) The composition of two functions A −
→B−
→ C is defined as g ◦ f : A → C,
g ◦ f (a) = g(f (a)). For example, h ◦ g(x) = x4 .
(a) If f and g are injective then g ◦ f is injective.
(b) If f andg are surjective then g ◦ f is surejective.
(c) If g ◦ f is injective then f is injective.
(d) If g ◦ f is surjective then g is surjective.
(11) A function f : A → B is bijective if f is both injective and surjective. The
function n is bijective.
(a) The identity function on A is idA : A → A defined by (surprise!)
idA (a) = a. idA is bijective.
g
←
−
(b) Suppose A −
→ B. If f ◦ g = idB and g ◦ f = idA then we say that
f
f and g are inverses of each other. Since idB is both injective and
surjective, f is surjective and g is injective. Since idA is both injective
and surjective, f is injctive and g is surjective. Thus if f and g are
inverses of each other, both are bijective.
(c) If f : A → B is bijective, then there exists a unique function g : B → A
which is the inverse of f . The inverse of n is p : R≥0 → R≥0 , p(x) =
√
x.
g
←
−
(12) Caution! Given A −
→ B, it is possible that f ◦ g = idB but g ◦ f 6= idA .
f
In that case we say f is a left inverse of g and g is a right inverse of f .
To check that one function is the inverse of another, you must check both
compositions f ◦ g and g ◦ f .
(a) Note in the case above that f ◦ g is both injective and surjective, so f
is surjective and g is injective. A left inverse is surjective and a right
inverse is injective.
(b) In fact, f has a right inverse if and only if f is surjective, and f has a
left inverse if and only if f is injective.
Department of Mathematics
San Francisco State University
San Francisco, CA 94132
E-mail address: [email protected]
URL: http://online.sfsu.edu/~meredith