Show that if both f and g are injective, then $g \circ f$ is also injective.

Show that if both f and g are surjective, then $g \circ f$ is also surjective.

Show, using a single counter example, that both of the converses to the results in part (a) and part (b) are false.

let s and t be in A and $s \ne t$     M1

since f is injective $f(s) \ne f(t)$     A1

since g is injective $g \circ f(s) \ne g \circ f(t)$     A1

hence $g \circ f$ is injective     AG

[3 marks]

let z be an element of C

we must find x in A such that $g \circ f(x) = z$     M1

since g is surjective, there is an element y in B such that $g(y) = z$     A1

since f is surjective, there is an element x in A such that $f(x) = y$     A1

thus $g \circ f(x) = g(y) = z$     R1

hence $g \circ f$ is surjective     AG

[4 marks]

converses: if $g \circ f$ is injective then g and f are injective

if $g \circ f$ is surjective then g and f are surjective     (A1)

    A2

Note: There will be many alternative counter-examples.

[3 marks]

This question was found difficult by a large number of candidates and no fully correct solutions were seen. A number of students made thought-through attempts to show it was surjective, but found more difficulty in showing it was injective. Very few were able to find a single counter example to show that the converses of the earlier results were false. Candidates struggled with the abstract nature of the question.

This question was found difficult by a large number of candidates and no fully correct solutions were seen. A number of students made thought-through attempts to show it was surjective, but found more difficulty in showing it was injective. Very few were able to find a single counter example to show that the converses of the earlier results were false. Candidates struggled with the abstract nature of the question.

This question was found difficult by a large number of candidates and no fully correct solutions were seen. A number of students made thought-through attempts to show it was surjective, but found more difficulty in showing it was injective. Very few were able to find a single counter example to show that the converses of the earlier results were false. Candidates struggled with the abstract nature of the question.