Problem Give an example where g is surjective and f is injec

Problem. Give an example where g is surjective and f is injective, but gf is not surjective and not injective.

We can think of this in terms of solving equations:

Given b B, how many solutions x A are there to the equation t(x) = b?

Solution of equations will be one of our main themes in linear algebra.

Solution

Case 1. Clearly f must be injective. Let A = {1,2}, B = {a,b}, C = {x}. f(1) = a, f(2) = b. g is not injective. g(a) =x, g(b) = x. Then g(f(1)) = g(a) = x, and g(f(2)) = g(b) = x, so g o f is not injective either. It follows that for g o f to be injective, g must be injective.

Case 2. Clearly g must be surjective. Let A = {1,2}, B = {a,b}, C = {x,y}. g(a) = x, g(b) = y. f is not surjective. f (1) = a, f(2) = a. Then g(f(1)) = x, g(f(2)) = x, nothing gets mapped to y in C, hence g o f is not surjective. It follows that for g o f to be surjective, f must be surjective.

Alternate:

The same example works for both. Let A = {1}, B = {1, 2}, C = {1}, and f : A B by f(1) = 1 and g : B C by g(1) = g(2) = 1. Then g f : A C is defined by (g f)(1) = 1. This map is a bijection from A = {1} to C = {1}, so is injective and surjective. However, g is not injective, since g(1) = g(2) = 1, and f is not surjective, since 2 6 f(A) = {1}.