Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 2F (Topics In Analysis)

LetThe function $h$ is continuous, withThe intermediate value theorem gives $c\in [0,1]$ such that $h(c)=0$, and hence $f(c)=c$. This proves the fixed-point property of a closed interval.

Now suppose $0,1\in f([0,1])$. For any $y\in [0,1]$, choose $a,b\in [0,1]$ with $f(a)=0$ and $f(b)=1$. Applying the intermediate value theorem to $f$ on the interval with endpoints $a,b$ gives a point $x$ with $f(x)=y$. Thus $f$ is a surjective function.

Let $h:A\to B$ be a homeomorphism, and let $f:B\to B$ be continuous. The conjugate mapis continuous. By hypothesis it has a fixed point $a\in A$. Thereforeand applying $h$ gives $f(h(a))=h(a)$. Hence every continuous self-map of $B$ has a fixed point. This is the homeomorphism invariance of the fixed-point property.

The Brouwer fixed-point theorem for the closed unit disc $\overline{D}\subset {\mathbb{R}}^2$ states that every continuous maphas a fixed point.

The closed unit disc has the fixed-point property by the Brouwer fixed-point theorem. The annulus admits the fixed-point-free rotation of an annulus, for examplewhich is continuous, maps the annulus to itself, and has no fixed point because the origin is not in the annulus. If the disc and annulus were homeomorphic, the homeomorphism invariance of the fixed-point property would give the annulus the fixed-point property, a contradiction. Therefore they are not homeomorphic.

No. TakeIt contains more than one point and has the fixed-point property of a closed interval, but it is not homeomorphic to the closed unit disc. Indeed, deleting any point $(t,0)$ with $0<t<1$ disconnects $B$, so such a point is a cut point. Deleting one point from a two-dimensional closed disc leaves a path-connected space. Since being a cut point is preserved by homeomorphisms, the two spaces cannot be homeomorphic. This gives the required counterexample.