Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 4 / 1E (Numbers and Sets)

An inverse function of $f:X\to Y$ is a function $h:Y\to X$ satisfying bothThese are two-sided inverse identities. If $f(x)=f(x^{\prime })$, apply $h$ to get $x=x^{\prime }$, so $f$ is injective. For every $y\in Y$, the identity $f(h(y))=y$ supplies a preimage, so $f$ is surjective. Thus $f$ is a bijection.

Conversely, if $f$ is a bijection, each $y\in Y$ has exactly one preimage. Define $h(y)$ to be that preimage. Then $f(h(y))=y$ and $h(f(x))=x$, giving both identities. Hence a function has a two-sided inverse exactly when it is a bijection. This also makes its inverse function unique.

Always true. If $(f\circ g)(x)=(f\circ g)(x^{\prime })$, injectivity of $f$ gives $g(x)=g(x^{\prime })$, and injectivity of $g$ gives $x=x^{\prime }$. Thus the function composition is injective. For any $y\in X$, since $f$ is surjective, choose $z$ with $f(z)=y$. Since $g$ is surjective, choose $x$ with $g(x)=z$. Then $(f\circ g)(x)=y$, proving that the composition is surjective. Consequently $f\circ g$ is a bijection; its inverse function is $g^{-1}\circ f^{-1}$.

Can be false. Take $X=\{0,1,2,\dots \}$ and defineThen $(f\circ g)(n)=n$, so their function composition is the identity function and hence a bijection. However, $g$ is not surjective, since zero has no preimage, while $f$ is not injective, since $f(0)=f(1)=0$.

The valid implications are only that $g$ must be injective and $f$ must be surjective: equality of two $g$ values implies equality after composition, and every output of the composition is an output of $f$. On a finite $X$, either of these properties implies a bijection, so an infinite set is essential to this counterexample.