Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 4 / 1D (Numbers and Sets)

False. Take $A=\{0\}$ and $B=C=\{0,1\}$, with $f(0)=0$ and $g={\mathrm{id}}_B$. Then $f$ is an injection and $g$ is a surjection, but the function composition $g\circ f$ has image of a function $\{0\}$, so it is not a surjection onto $C$.

True. The function composition $j=g\circ f$ is an injection: $j(a)=j(a^{\prime })$ first implies $f(a)=f(a^{\prime })$ by the injectivity of $g$, then $a=a^{\prime }$ by the injectivity of $f$. Fix one $a_0\in A$, using the given nonemptiness, and defineThe first case is well defined because $j$ is an injection. For every $a\in A$, $h(j(a))=a$. Thus this left inverse satisfies $\boxed{h\circ g\circ f={\mathrm{id}}_A}$, the identity function on $A$.

False. Take $A=\{0,1\}$, $B=\{0\}$ and the constant function $f(a)=0$. For the subsets $X=\{0\}$ and $Y=\{1\}$, the intersection $X\cap Y$ is empty, so $f(X\cap Y)=\varnothing$, whereas the intersection of their images of a function is $f(X)\cap f(Y)=\{0\}$. In general the inclusion $f(X\cap Y)\subseteq f(X)\cap f(Y)$ holds, but a common image value can arise from different points of $X$ and $Y$.

True. For each $a\in A$, the definition of preimage givesThus $\boxed{f^{-1}(Z\cap W)=f^{-1}(Z)\cap f^{-1}(W)}$. Unlike the corresponding statement for images of a function, this intersection identity requires no injectivity.