Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 7E (Numbers and Sets)

If $f$ is injective and $x\in f^{-1}(f(A))$, then $f(x)=f(a)$ for some $a\in A$, so injectivity gives $x=a\in A$. The reverse inclusion always holds; hence $f^{-1}(f(A))=A$.

Conversely, if (ii) holds and $f(x)=f(y)$, then $y\in f^{-1}(f(\{x\}))=\{x\}$, so $x=y$. Thus (i) and (ii) are equivalent.

Assume $f$ is injective. The inclusion $f(A\cap B)\subseteq f(A)\cap f(B)$ holds for every function. If $y$ belongs to the right-hand side, write $y=f(a)=f(b)$ with $a\in A$ and $b\in B$. Injectivity gives $a=b\in A\cap B$, proving equality.

Conversely, suppose (iii) holds and $f(x)=f(y)$. If $x\ne y$, take $A=\{x\}$ and $B=\{y\}$. Then $f(A)\cap f(B)$ is nonempty whereas $f(A\cap B)=f(\varnothing )=\varnothing$, a contradiction. Hence $f$ is injective.

Parts (i) and (ii) prove (i)$\iff$(ii), while the two directions in part (ii) prove (i)$\iff$(iii). Therefore all three statements are equivalent.

DefineCertainly $f(B)\subseteq B$. If $y\in B$, then $y\in f(X)$ and has a unique predecessor $x$ because $f$ is injective. For every $n$, writing $y=f^{n+1}(z)$ and using uniqueness gives $x=f^n(z)$; hence $x\in B$ and $y\in f(B)$. Thus $f(B)=B$.

If $f(a)\in B$, its unique predecessor must lie in $B$, so $a\in B$; consequently $f(A)\subseteq A$. If a point belonged to every $f^n(A)$ for $n\ge 1$, it would belong to $B$ and also to $f(A)\subseteq A$, which is impossible. Therefore