Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 1 / c

The function is injective and surjective, with inverse function $f^{-1}(y)=1-y$. Equality $1-x=1-x^{\prime }$ forces $x=x^{\prime }$. For any $y\in (0,1)$, the input $x=1-y$ also lies in $(0,1)$ and satisfies $f(x)=y$. Applying the function twice returns the original input.

It is neither injective nor surjective onto $\mathbb{R}$, so it has no inverse function on the stated domain and codomain. The distinct inputs $0$ and $\pi /3$ both lie in the domain and both give zero. All values lie in $[-1,1]$, so, for example, the real number $2$ has no preimage.

It is surjective but not injective; there is no global inverse function. Noninjectivity follows fromFor surjectivity, restrict to $x<1$, whereThis restriction is a continuous function, tends to $-\infty$ as $x\to -\infty$ and to $+\infty$ as $x\to 1^{-}$. The intermediate value theorem therefore supplies a preimage for every real number. The additional branch $x>1$ destroys global injectivity but cannot destroy surjectivity.