Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 21F / iii / Solution

Yes. Let $b$ be the vertex of $({\Delta }^3{)}^{\prime }$ corresponding to the whole tetrahedron. Its link in $M$ is the 1-skeleton of the barycentric subdivision of $\partial {\Delta }^3$. This connected graph has one vertex for each of the 14 nonempty proper faces and 36 edges, so

No other point has this local homology rank. At the other vertices, corresponding respectively to faces of size $1,2,3$, the first Betti numbers of the links are $6,4,6$. A point in the interior of an edge is incident with at most six triangles, so its local $H_2$ has rank at most five; a point in a triangle interior has local $H_2\cong \mathbb{Z}$.

A homeomorphism preserves local homology, so it must send the unique point with local $H_2$ of rank $23$ to itself. Therefore every self-homeomorphism fixes $b$, which is the fixed barycentre of the barycentric tetrahedral two-skeleton.

Solved by gpt-5.6-sol high.