Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 21H (Algebraic Topology)

For every nonempty simplex $\sigma$ of $K$, let $b_{\sigma }$ be its barycentre. The first barycentric subdivision $K^{\prime }$ has the $b_{\sigma }$ as vertices, andis a simplex exactly when, after reordering, ${\sigma }_0⊊\cdots ⊊{\sigma }_q$. Its realization is the same polyhedron as that of $K$. The iterated barycentric subdivision is defined by

The mesh of a simplicial complex is the largest diameter of one of its realized simplices, or the supremum when $K$ is not finite:If $\dim K=n$, thenIterating gives

A simplicial map $g:K\to L$ is a simplicial approximation to $f:|K|\to |L|$ whenfor every vertex $v$, where $\mathrm{st}$ denotes the open star in a simplicial complex.

Fix $x$ in a simplex with vertices $v_0,\dots ,v_q$, retaining only vertices with positive barycentric coordinate at $x$. Then $x\in {\mathrm{st}}_K(v_i)$ for every $i$, so $f(x)$ lies in every ${\mathrm{st}}_L(g(v_i))$. The vertices $g(v_i)$ and the vertices carrying the barycentric coordinates of $f(x)$ consequently span a common simplex of $L$. Both $f(x)$ and $|g|(x)$ lie in its convex realization. The straight-line homotopy from a simplicial approximationis therefore well-defined and continuous, with endpoints $f$ and $|g|$.

The Simplicial approximation theorem states that if $K$ is a finite simplicial complex, $L$ is a simplicial complex, and $f:|K|\to |L|$ is continuous, then for some $r\ge 0$ there is a simplicial approximation $g:K^{(r)}\to L$ to $f$.

The target open stars form an open cover of $|L|$, so their inverse images under $f$ form an open cover of the compact metric space $|K|$. The Lebesgue number lemma supplies $\delta >0$ such that every subset of diameter below $\delta$ lies in one inverse image. Choose $r$ so large that every vertex star in $K^{(r)}$ has diameter below $\delta$; this is possible because its diameter is at most twice $\mu (K^{(r)})$ and the mesh tends to zero.

For each vertex $v$ choose a vertex $w(v)$ of $L$ such thatIf $v_0,\dots ,v_q$ span a simplex, a point in its relative interior belongs to all their open stars. Its image belongs to all the open stars of $w(v_0),\dots ,w(v_q)$, so these target vertices span a simplex. Thus $v\mapsto w(v)$ extends to a simplicial map $g$, and its defining inclusion says exactly that $g$ approximates $f$. This proves the theorem.

Triangulate $S^n$ and $S^m$. By the Simplicial approximation theorem, after subdividing the domain, a continuous map $f:S^n\to S^m$ is homotopic to the realization of a simplicial map $g$. Since $n<m$, the image of $|g|$ lies in the $n$-skeleton of the target triangulation and therefore misses an interior point of every $m$-simplex. A sphere with one point removed is homeomorphic to the Euclidean space ${\mathbb{R}}^m$, which is contractible. Hence $|g|$, and therefore $f$, is homotopic to a constant map. Thus