Chapter 4 Flashcards
Explain the notation:
f : A –> B, a |–> b
The first part A –> B signifies that f transforms all elements from the input set A (domain), into elements of the output set B (codomain).
The second part a |–> b indicates what happens at an element level.
How is the range denoted?
f(A)
Define a function.
For f : A –> B, a |–> b, every a ∈ A gives just one b = f(a).
Denote the set of all pre-images of b. Give another name.
f^(<-) (b)
Inverse image of b for f
What is inverse image of b for f if b is outside the range?
Null.
Define equality of functions.
Two functions f : A –> B and g : A –> B are equal if f(a) = g(a) for all a ∈ A.
Define injective. Give another name.
f : A –> B is injective if
for all a1, a2 ∈ A, if f(a1) = f(a2) then a1 = a2
One to one
What is equivalent to saying for all b ∈ B, if there exists a ∈ A s.t. b = f(a), then a is the unique pre-image of b?
f is injective
Define surjective. Give another name.
f: A –> B is surjective for every b ∈ B there exists a ∈ A s.t. f(a) = b.
Or more simply: B = f(A)
T or F. A function can be bijective but not injective.
False.
T or F. A function can be injective but not surjective.
True.
T or F. A function that is surjective and injective has an inverse.
True.
T or F. Every bijective function has an inverse.
True.
Let A, B and C be sets, and let f : A –> B and g : B –> C be two functions. The composition of f and g, denoted g . f is defined…
g . f : A –> C, a |–> g(f(a)), for all a ∈ A.
T or F. The composition f.g is always equal to the composition g.f?
False.
T or F. The composition of two bijective, injective or surjective functions are consequently bijective, injective or surjective.
True.
Define the identity map on A.
idA : A –> A, a |–> a
Define whether a function is inveritble.
f: A –> B is invertible if there exists g : B –> A
s.t. g(f(a)) = a for all a ∈ A
and f(g(b)) = b for all b ∈ B
T or F. If a function has an inverse it is unique.
True.
T or F. If f: A -> B is invertible, then f^-1 : B –> A is invertible and (f^-1)^-1 = f.
True.
If f: A –> B and g: B –> C are both invertible what can be said about g . f : A –> C
g.f is invertible
(g.f)^-1 = f^-1.g^-1
T or F. Let f: A –> B. If f is bijective |A| == |B|.
True.
