47.9k views
4 votes
Find sets X, Y, and Z and functions f : X → Y and g : Y → Z such that (a) g is injective, but g ◦ f is not injective.

(b) g ◦ f is surjective, but f is not surjective.

(c) g is surjective, but g ◦ f is not surjective.

(d) f is surjective, but g ◦ f is not surjective.

(e) f is injective, but g ◦ f is not injective.

(f) g ◦ f is injective but g is not injective.

(g) g is injective, but g ◦ f is not injective.

User Baretta
by
7.8k points

1 Answer

2 votes

Final answer:

The questions involve creating examples of functions and sets to demonstrate specific injective and surjective properties with function compositions. Scenarios (c), (d), (e), and (f) are impossible given the constraints. Scenario (a) is possible with g injective and g ◦ f not injective, and (b) has a surjective composition with a non-surjective f.

Step-by-step explanation:

The question concerns concepts related to sets and functions, specifically focusing on the properties of injectivity (one-to-one function) and surjectivity (onto function) in the context of function composition.

Example Solutions

a) g is injective, but g ◦ f is not injective:
Let X = {1,2}, Y = {a,b}, Z = {α, β}. Define f: X → Y by f(1) = a and f(2) = a (not injective), and g: Y → Z by g(a) = α and g(b) = β (injective). g ◦ f is not injective because g(f(1)) = g(f(2)) = α.

b) g ◦ f is surjective, but f is not surjective:
Keep X, Y, and Z as before with the same functions. g ◦ f is surjective since every element in Z has a pre-image in X. However, f is not surjective since 'b' in Y has no pre-image in X.

c) g is surjective, but g ◦ f is not surjective:
This scenario is impossible under the given conditions as the surjectivity of g ensures that for every element in Z, there is at least one element in Y, and since every element in X maps to Y via f, g ◦ f would also be surjective.

d) f is surjective, but g ◦ f is not surjective:
This scenario is impossible as the surjectivity of f would imply that g ◦ f must cover all elements of Z if g is a function from Y to Z.

e) f is injective, but g ◦ f is not injective:
This scenario is also impossible as the injectivity of f would be maintained through the composition, thus if g ◦ f is not injective, then g must not be injective.

f) g ◦ f is injective but g is not injective:
As with (e), this is impossible under the given conditions as the non-injectivity of g would lead to g ◦ f also being non-injective.

g) g is injective, but g ◦ f is not injective:
This is the same as case (a) and it is possible as described above in (a).

User Fraff
by
7.9k points