Prove that: 

(i)     \(f\) is an injection,

(ii)     \(g\) is a surjection.

Given that \(X = {\mathbb{R}^ + } \cup \{ 0\} \) and \(Y = \mathbb{R}\), choose a suitable pair of functions \(f\) and \(g\) to show that \(g\) is not necessarily a bijection.

(i)     to test injectivity, suppose \(f({x_1}) = f({x_2})\)     M1

apply \(g\) to both sides \(g\left( {f({x_1})} \right) = g\left( {f({x_2})} \right)\)     M1

\( \Rightarrow {x_1} = {x_2}\)     A1

so \(f\) is injective     AG

Note:     Do not accept arguments based on “\(f\) has an inverse”.

(ii)     to test surjectivity, suppose \(x \in X\)     M1

define \(y = f(x)\)     M1

then \(g(y) = g\left( {f(x)} \right) = x\)     A1

so \(g\) is surjective     AG

[6 marks]

choose, for example, \(f(x) = \sqrt x \) and \(g(y) = {y^2}\)     A1

then \(g \circ f(x) = {\left( {\sqrt x } \right)^2} = x\)     A1

the function \(g\) is not injective as \(g(x) = g( - x)\)     R1

[3 marks]

Total [9 marks]

Those candidates who formulated the questions in terms of the basic definitions of injectivity and surjectivity were usually sucessful. Otherwise, verbal attempts such as '\(f{\text{ is one - to - one }} \Rightarrow f{\text{ is injective}}\)' or '\(g\) is surjective because its range equals its codomain', received no credit. Some candidates made the false assumption that \(f\) and \(g\) were mutual inverses.

Few candidates gave completely satisfactory answers. Some gave functions satisfying the mutual identity but either not defined on the given sets or for which \(g\) was actually a bijection.