If f: A ➡B and g:B ➡ C are injective
function, then gof: A ➡C is an injective
function. Prove or disprove.
Answers
Step-by-step explanation:
1. Show that if f : A −→ B and g : B −→ C are functions such that g ◦ f is injective, then f is
injective.
Solution: Assume that g ◦ f is injective. We want to show that f is injective, so suppose that
a, a0 ∈ A are such that f(a) = f(a
0
) (we will be done if we can show that a = a
0
). Applying g
to both sides of the equation we obtain that g(f(a)) = g(f(a
0
)). But g(f(x)) = (g ◦ f)(x) and so
(g ◦ f)(a) = (g ◦ f)(a
0
). This implies that a = a
0 using the assumption that g ◦ f is injective. But
then we have just shown that f is injective too.
Remark: The thing I would have most liked to see on people’s solutions is an explanation of
how one goes from f(a) = f(a
0
) to g(f(a)) = g(f(a
0
)), in other words saying that they are applying
g to both sides. This is one of my favorite problems to put on midterms and finals.
2. Give an example of functions f : A −→ B and g : B −→ C such that g ◦ f is injective but g is not
injective.
Solution: There are lots of correct answers. You can set A = {1}, B = {2, 3} and C = {4}.
Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. Then g is not
injective (since both 2, 3 7→ 4) but g ◦ f is injective.
Here’s another correct answer using more familiar functions.
Let f : R≥0 −→ R be given by f(x) = √
x. Let g : R −→ R be given by g(x) = x
2
. Then g is not
injective (since g(1) = g(−1)) but g ◦ f : R≥0 −→ R is injective since it sends x 7→ x.
Remark: Lots of groups did some variant of the second example. I took off points if they didn’t
specify the domain and codomain though. Note that the codomain of f must equal the domain of
g for g ◦ f to make sense.
3. Suppose that f : A −→ B and g : B −→ C are functions and that g ◦ f is surjective. Is it true
that f must be surjective? Is it true that g must be surjective? Justify your answers with either a
counterexample or a proof.
Solution: There are two questions in this problem.
Must f be surjective? The answer is no. Indeed, let A = {1}, B = {2, 3} and C = {4}.
Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. We see that
g ◦ f : {1} −→ {4} is surjective (since 1 7→ 4) but f is certainly not surjective.
Must g be surjective? The answer is yes, here’s the proof. Suppose that c ∈ C is arbitrary (we
must find b ∈ B so that g(b) = c, at which point we will be done). Since g ◦ f is surjective, for the
c we have already fixed, there exists some a ∈ A such that c = (g ◦ f)(a) = g(f(a)). Let b := f(a).
Then g(b) = g(f(a)) = c and we have found our desired b.
Remark: It is good to compare the answer to this problem to the answer to the two problems
on the previous page.
The part of this problem most groups had the most issue with was the second. Everyone should
be comfortable with carefully proving a function is surjective by the time we get to the midterm.