Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 20I / Solution

Suppose first that $F:D^n\to X$ extends $f$. Regarding $S^{n-1}$ as the boundary of the unit disc,is a homotopy from $f$ to the constant map with value $F(0)$. Thus $f$ is null-homotopic. Conversely, if $H:S^{n-1}\times [0,1]\to X$ is a homotopy from $f$ to a constant, then $H$ is constant on $S^{n-1}\times \{1\}$ and hence descends to the quotientwhich is the cone on $S^{n-1}$ and is homeomorphic to $D^n$. The descended map extends $f$. This proves the extension-null-homotopy criterion for a sphere.

A universal cover of $X$ is a covering map $p:\tilde{X}\to X$ whose total space is path-connected and simply connected. Let $p_1:({\tilde{X}}_1,{\tilde{x}}_1)\to (X,x)$ and $p_2:({\tilde{X}}_2,{\tilde{x}}_2)\to (X,x)$ be two universal covers. The lifting criterion for a covering space applies becauseso $p_1$ has a based lift $h:{\tilde{X}}_1\to {\tilde{X}}_2$ satisfying $p_{2}h=p_1$. Similarly there is a based lift $k:{\tilde{X}}_2\to {\tilde{X}}_1$ of $p_2$. Both $kh$ and the identity are lifts of $p_1$ that agree at the base point, so uniqueness of lifts gives $kh=1$. Likewise $hk=1$. Hence $h$ is a homeomorphism. This is the uniqueness of a universal covering space.

Now let $p:\tilde{X}\to X$ be universal with $\tilde{X}$ contractible. If an extension $F:|K|\to X$ exists, functoriality of the fundamental group givesThus the required factorization holds with $\Phi =F_{\ast }$.

Conversely, suppose $f_{\ast }=\Phi i_{\ast }$. We extend over the simplices of $K$ by dimension. The boundary of every 2-simplex is a loop in $|K^1|$ that becomes null-homotopic in $|K|$, so its class lies in $\ker i_{\ast }$. The factorization givesHence $f{|}_{\partial \sigma }$ is null-homotopic in $X$ and extends over the disc $|\sigma |$ by the first part. Since distinct 2-simplices meet along the already fixed 1-skeleton, these extensions combine to a map on $|K^2|$.

Inductively suppose the map is defined on $|K^{m-1}|$ for $m\ge 3$. For an $m$-simplex $\sigma$, its boundary is $S^{m-1}$, which is simply connected. The boundary map therefore lifts through $p$ by the covering-space lifting criterion. Its lift into the contractible space $\tilde{X}$ is null-homotopic, so the boundary map itself is null-homotopic and extends over $|\sigma |$. Extending over every $m$-simplex and then every skeleton constructs $F:|K|\to X$. The weak topology on a simplicial complex makes the cellwise map continuous. This proves the extension criterion into a space with contractible universal cover.

Solved by gpt-5.6-sol high.