Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 21F (Algebraic Topology)

The universal coefficient theorem for homology and flatness of $\mathbb{Q}$ giveThus, ifwith $T_i$ torsion, then $H_i(X;\mathbb{Q})\cong {\mathbb{Q}}^{b_i}$.

For a finite triangulable space, its Euler characteristic isIf $K$ is a finite triangulation and $f_i$ is its number of $i$-simplices, then the Euler-Poincare formula isTo prove it over $\mathbb{Q}$, write $Z_i=\ker {\partial }_i$ and $B_i=\mathrm{im}{\partial }_{i+1}$. Rank-nullity and $H_i=Z_i/B_i$ giveAfter multiplying by $(-1{)}^i$ and summing, the two boundary sums cancel, leaving the claimed equality.

The standard simplex ${\Delta }^3$ is contractible, so

The simplices in a barycentric subdivision are strict chains of nonempty faces. A tetrahedron has $15$ nonempty faces, so $f_0=15$. Counting strict two-face chains givesA strict three-face chain amounts to choosing a terminal face and partitioning its vertices into three nonempty ordered blocks. Hencewhere $S(-,-)$ denotes a Stirling number of the second kind. ThereforeEquivalently, the full subdivision has $4!=24$ tetrahedra and Euler characteristic one, so deleting its three-dimensional simplices from the alternating count gives $1+24=25$. This is the f-vector computation for the barycentric subdivision of a tetrahedron.

The complex $M$ is connected, so $H_0(M)\cong \mathbb{Z}$. Adding the 24 three-simplices to $M$ produces the contractible complex $({\Delta }^3{)}^{\prime }$. Adding three-simplices does not alter $H_1$, henceAlso $H_i(M)=0$ for $i>2$ because $M$ is two-dimensional.

The group $H_2(M)=\ker ({\partial }_2)$ is a subgroup of the free group $C_2(M)$ and is therefore free abelian. If its rank is $b_2$, the Euler characteristic from part (i) givesso $b_2=24$. Consequently

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.