Function Composition

Question

For gof to be one-to-one, f must be one-to-one but reverse is not true.

Is the above statement correct?

According to me both f and g must be one-to-one for gof to be one-to-one.

I took below examples:

(i) f is one-to-one and g is onto

Let A = {1, 2, 3}, B = {a, b, c} and C = {α, β}.

Given f: A -> B where f(1) = a, f(2) = b, f(3) = c and g: B -> C where g(a) = α, g(b) = α, g(c) = β.

Then, gof: A -> C which gives g(f(1)) = g(a) = α, g(f(2)) = g(b) = α and g(f(3)) = g(c) = β.

Here, g(f(1)) and g(f(2)) both gives α which implies that gof is not one-to-one.

(ii) f is onto and g is one-to-one

Let A = {1, 2, 3}, B = {a, b} and C = {α, β}.

Given f: A -> B where f(1) = a, f(2) = a, f(3) = b and g: B -> C where g(a) = α, g(b) = α.

Then, gof: A -> C which gives g(f(1)) = g(a) = α, g(f(2)) = g(a) = α and g(f(3)) = g(b) = β.

Here, also g(f(1)) and g(f(2)) both gives α which implies that gof is not one-to-one.

(ii) f is one-to-one and g is one-to-one

Let A = {1, 2}, B = {a, b} and C = {α, β}.

Given f: A -> B where f(1) = a, f(2) = b and g: B -> C where g(a) = α, g(b) = α.

Then, gof: A -> C which gives g(f(1)) = g(a) = α and g(f(2)) = g(b) = β.

Here, g(f(1)) and g(f(2)) both gives α which implies that gof is one-to-one.

Are above examples correct? If yes, please explain.

Answer 1

listen these all are one way true

1) if gof is one-one then f has to one-one

2) if gof is onto then g has to onto

3) if f is one-one and g is one-one then gof is one-one

4) if f is onto and g is onto then gof is onto