Let \(S_n\) be the circle of radius \(n\) in \(\mathbb{R}^2\) whose center is at the point \((n, 0)\). Let \(Y\) be the subspace of \(\mathbb{R}^2\) that is the union of these circles; let \(p\) be their common point.
Show that \(Y\) is not homeomorphic to a countably infinite wedge \(X\) of circles, nor to the space of Example 1.
Show, however, that \(\pi_1(Y, p)\) is a free group with \(\ts{[f_n]}\) as a system of free generators, where \(f_n\) is a loop representing a generator of \(\pi_1(S_n, p)\).
Solution
Let \(p=(0,0)\). The circle \(S_n\) has equation \[ x^2+y^2=2nx. \]
- The countably infinite wedge \(\bigvee_{n\ge1}S^1\), with the wedge topology, is not first countable at its wedge point (Exercise 4 of this section). The space \(Y\subset\mathbb R^2\) is metrizable, hence first countable. Therefore these spaces are not homeomorphic.
Example 1 of the section is the infinite earring formed by circles of radius \(1/n\). That space is bounded and closed in \(\mathbb R^2\), hence compact. By contrast, \(Y\) is unbounded: \((2n,0)\in S_n\) for every \(n\). Thus \(Y\) is not compact and cannot be homeomorphic to the infinite earring.
- For \(N\ge1\), put \[ Y_N=S_1\cup\cdots\cup S_N. \] This is a finite wedge of circles, so \[ \pi_1(Y_N,p)\cong F(f_1,\dots,f_N). \] The inclusions \(Y_N\hookrightarrow Y_{N+1}\) send these generators to the corresponding generators.
We claim every loop in \(Y\) is homotopic rel basepoint to a loop in some \(Y_N\). Let \(g:I\to Y\) be a loop. Its image is compact, hence bounded; choose \(R>0\) with \[ g(I)\subset B_R(p). \] Choose \(N>R\). For \(n>N\), the intersection \(S_n\cap B_R(p)\) is a proper arc-neighborhood of \(p\) in \(S_n\), hence contractible to \(p\) within \(S_n\cap B_R(p)\). Indeed, on \(S_n\) the radial distance \(r\) from \(p\) satisfies \(r=2n\cos\theta\), so the condition \(r<R< n\) cuts out only the two short branches meeting at \(p\), together forming an arc.
Consequently each excursion of \(g\) into \(S_n-\{p\}\) for \(n>N\) can be contracted to \(p\). These contractions can be performed simultaneously. To check continuity at an accumulation time of excursions, use uniform continuity of \(g\): the parameter intervals of any infinite family of pairwise disjoint excursions have lengths tending to zero, hence the diameters of their images tend to zero; choose the arc contractions so that the distance to \(p\) never increases. Thus the resulting homotopy is continuous and produces a loop with image in \(Y_N\).
It follows that \[ \pi_1(Y,p)=\bigcup_{N\ge1}\operatorname{im}\bigl(\pi_1(Y_N,p)\to\pi_1(Y,p)\bigr). \] It remains to see that no new relations appear. For each \(N\), define a retraction \[ r_N:Y\to Y_N \] by the identity on \(Y_N\) and by sending every circle \(S_n\), \(n>N\), to \(p\). This is continuous: away from \(p\) only one circle is locally present, while at \(p\), every neighborhood of \(p\) in \(Y_N\) has preimage equal to that neighborhood together with all collapsed circles, hence contains a neighborhood of \(p\) in \(Y\). Therefore the inclusion-induced map \[ \pi_1(Y_N,p)\to\pi_1(Y,p) \] is injective, since \((r_N)_*\) is a left inverse.
Hence \(\pi_1(Y,p)\) is the increasing union \[ F(f_1)\subset F(f_1,f_2)\subset\cdots, \] which is the free group on the countable set \(\{[f_n]:n\ge1\}\). Thus \[ \boxed{\pi_1(Y,p)\cong F([f_1],[f_2],\dots).} \]