Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 7E (Numbers and Sets)

True. Take $C=f(A)$ and define $g(f(a))=a$, which is well-defined and injective because $f$ is injective.

False. A constant surjection from a two-element set to a singleton followed by the singleton’s inclusion is not injective.

False. Include a singleton into a two-element set and then map both elements onto a singleton; the composite need not cover a larger target in analogous examples. Concretely $A=\{1\}$, $B=\{1,2\}$, $C=\{1,2\}$, $f(1)=1$, and let $g$ swap/surject with $g(1)=g(2)=1$ after taking $C$ singleton; then the claimed implication fails whenever an element of $C$ is reached only outside $f(A)$.

True. If $f(a_1)=f(a_2)$ with $a_1\ne a_2$, then $(g\circ f)(a_1)=(g\circ f)(a_2)$.

Put $|B|=m$. There are $\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)m!=(m+1)!m/2$ surjections and $(m+1)!$ injections. For $m\ge 3$ the domain is larger, so $\Psi$ cannot be injective, regardless of the choices. For $m=1$ there is only one surjection, so it is injective. For $m=2$ both sets have size $6$; injectivity depends on the representative choices and, if achieved, is equivalent to bijectivity.