Past exam of the topology and groups course of the University of Oxford / 2023 / 3 / c / Solution

Use a topological graph consisting of the ray $[0,\infty )$ with vertices $v_n=n$ and one oriented graph loop ${\ell }_n$ attached at each $v_n$, for $n=0,1,2,\dots$. Set the base point to $b=v_0$. This is a connected, locally finite graph; it is deliberately infinite.

Define continuous maps $f,g:X\to X$ byHere each graph loop is parametrized by $s\in [0,1]$ with its endpoints at the attaching vertex. The formulas agree at every vertex, and their restrictions to every closed edge are continuous. The weak topology of a CW complex therefore makes $f,g$ continuous maps. Directly on the ray and on each graph loop,

We check the two fundamental group maps at exactly the specified base points. The ray is a spanning tree. At $b=v_0$ a free basis of a group consists of $a_n$, obtained by going along the ray to $v_n$, following ${\ell }_n$, and returning. At $f(b)=v_1$, use the corresponding free basis of a group $c_n$ with connecting paths starting at $v_1$; in particular $c_0$ travels backwards along the first edge before following ${\ell }_0$. The standard spanning tree description applies to this infinite topological graph as well: every loop and every homotopy between based loops has compact image, so lies in a finite subcomplex.

The induced group homomorphisms areThus $f_{\ast }$ misses $c_0$: words in $c_1,c_2,\dots$ cannot equal $c_0$ by the normal form for free groups and free product of groups. Consequently $f_{\ast }$ is not surjective. Meanwhile $c_0$ is a nontrivial free group element killed by $g_{\ast }$, so $g_{\ast }$ is not injective. Neither induced map is a group isomorphism, while functoriality of the fundamental group correctly gives $g_{\ast }{f}_{\ast }=\mathrm{id}$.