Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 4 / 7D (Numbers and Sets)

Injective means $f(x)=f(y)$ implies $x=y$. It need not imply surjective on an infinite set: $f(n)=n+1$ is injective from $\mathbb{N}$ to itself but misses its least element.

An element of the left side is $f(s)$ for some $s\in S$ with $f(s)\in T$, exactly an element of $T\cap f(S)$. The second equality need not hold, even for a surjection: map $\{1,2\}$ to a singleton and take $S=\{1\}$, $S^{\prime }=\{2\}$. Then the image of the intersection is empty but the intersection of the images is not.

True. If $p(a,a^{\prime })=p(b,b^{\prime })$, then $a^{\prime }=b^{\prime }$, so $f(a)=g(a^{\prime })=g(b^{\prime })=f(b)$. Injectivity of $f$ gives $a=b$.

True. Given $a^{\prime }\in A^{\prime }$, surjectivity supplies $a\in A$ with $f(a)=g(a^{\prime })$. Then $(a,a^{\prime })$ lies in the fibre product and maps to $a^{\prime }$.