In this problem we use the notions of suspension, reduced suspension, cone, and mapping cone defined in Chapter 0. Let \(X\) be the subspace of \(\mathbb{R}\) consisting of the sequence \(1, {^1/_2}, {^1/_3}, {^1/_4}, \cdots\) together with its limit point 0.
For the suspension \(SX\), show that \(\pi_1(SX)\) is free on a countably infinite set of generators, and deduce that \(\pi_1(SX)\) is countable. In contrast to this, the reduced suspension \(\Sigma X\), obtained from \(SX\) by collapsing the segment \(\{0\} \times I\) to a point, is the shrinking wedge of circles in Example 1.25, with an uncountable fundamental group.
Let \(C\) be the mapping cone of the quotient map \(SX \to \Sigma X\). Show that \(\pi_1(C)\) is uncountable by constructing a homomorphism from \(\pi_1(C)\) onto \(\Pi_\infty \mathbb{Z} / \oplus_\infty \mathbb{Z}\). Note that \(C\) is the reduced suspension of the cone \(CX\). Thus the reduced suspension of a contractible space need not be contractible, unlike the unreduced suspension.
Solution
Write \[ x_n=\frac1n\quad(n\ge1), \qquad x_0=0. \] In \(SX\), let \(E_n\) be the suspension arc coming from \(\{x_n\}\times I\), and let \(S,N\) be the two suspension vertices. Thus \(SX\) is the union of the countably many arcs \(E_n\), all with endpoints \(S,N\), with \(E_n\) accumulating onto \(E_0\).
1. Any loop in \(SX\) crosses from \(S\) to \(N\), or from \(N\) to \(S\), only finitely many times.
The suspension coordinate defines a continuous function \[ h:SX\to I, \qquad h(S)=0,\quad h(N)=1,\quad h([x,t])=t. \] For a loop \(\gamma:I\to SX\), the composite \(h\gamma:I\to I\) is uniformly continuous. Choose \(\delta>0\) such that \[ |s-t|<\delta \quad\Longrightarrow\quad |h\gamma(s)-h\gamma(t)|<\frac12. \] Any subinterval on which \(\gamma\) goes from \(S\) to \(N\) or from \(N\) to \(S\) has endpoints whose \(h\gamma\)-values differ by \(1\), hence must have length at least \(\delta\). There can be only finitely many pairwise disjoint such subintervals in \(I\).
2. Every loop in \(SX\) is homotopic to a finite edge loop using only finitely many arcs \(E_n\).
Away from \(S\) and \(N\) one has \[ SX\setminus\{S,N\}\cong X\times(0,1). \] Since \(X\) is totally disconnected, each path component of this complement is one open arc \[ E_n\setminus\{S,N\}. \] Thus between successive visits to \(S\) or \(N\), the loop stays in a single \(E_n\).
1.Hence the loop is homotopic to a finite concatenation of full traversals of finitely many \(E_n\)’s.
3. For \(n\ge1\), let \(a_n\) be the loop that travels from \(S\) to \(N\) along \(E_n\) and returns from \(N\) to \(S\) along \(E_0\).Then the classes \([a_n]\) generate \(\pi_1(SX)\).
By 2, every loop is a finite edge loop. Insert or delete occurrences of \(E_0\bar E_0\) between successive traversals to rewrite each full traversal of \(E_n\) relative to the reference edge \(E_0\). The resulting loop is a finite word in the \(a_n\) and their inverses.
4. The classes \([a_n]\) satisfy no nontrivial finite relation.
Fix \(m\) and let \[ K_m=E_0\cup E_1\cup\cdots\cup E_m. \] Define \[ r_m:X\to\{x_0,x_1,\dots,x_m\} \] by fixing \(x_0,x_1,\dots,x_m\) and sending every \(x_n\) with \(n>m\) to \(x_0\). This map is continuous: the only nonisolated point of \(X\) is \(x_0\), and the entire tail is sent to \(x_0\). Suspending \(r_m\) gives a retraction \[ R_m:SX\to K_m. \]
The finite graph \(K_m\) consists of \(m+1\) parallel edges between \(S\) and \(N\), so \[ \pi_1(K_m)\cong F(a_1,\dots,a_m). \] If a reduced word in \(a_1,\dots,a_m\) were nullhomotopic in \(SX\), applying \((R_m)_*\) would make the same reduced word trivial in this free group, which is impossible. Thus there is no nontrivial finite relation among the \([a_n]\).
5. Therefore\[ \boxed{ \pi_1(SX)\cong F(a_1,a_2,\dots), } \]the free group on countably infinitely many generators; in particular, \(\pi_1(SX)\) is countable.
4.A word in a countable alphabet has finite length, and the set of all finite words in a countable alphabet is countable.
6. Collapsing \(E_0\) gives the reduced suspension \(\Sigma X\), the shrinking wedge of circles, and there is a surjective winding homomorphism\[ \rho:\pi_1(\Sigma X)\longrightarrow\prod_{n=1}^{\infty}\mathbb Z. \]
After \(E_0\) is collapsed, each \(E_n\cup E_0\) becomes a circle \(C_n\), and the diameters of these circles shrink to the common basepoint. This is the shrinking wedge of circles of Example 1.25. For each \(n\), collapse all circles except \(C_n\) to the basepoint to obtain a retraction \[ s_n:\Sigma X\to C_n. \] The product of the induced winding-number homomorphisms is \(\rho\).
Given any integer sequence \((k_n)\), traverse \(C_n\) exactly \(k_n\) times during a sequence of time intervals accumulating at \(1\). Since the circles shrink to the basepoint, the resulting infinite concatenation is continuous at the limiting time. Its image under \(\rho\) is \((k_n)\), so \(\rho\) is surjective.
7. Let\[ q:SX\to\Sigma X \]be the quotient map.Then\[ \rho\bigl(q_*(\pi_1(SX))\bigr) = \bigoplus_{n=1}^{\infty}\mathbb Z. \]
By 5, \(\pi_1(SX)\) is generated by the loops \(a_n\). The quotient \(q\) sends \(a_n\) to the standard loop once around \(C_n\). Therefore \[ \rho(q_*[a_n])=e_n, \] the \(n\)th standard basis vector. Every element of \(\pi_1(SX)\) is a finite word, so its coordinate vector has finite support. Conversely every finite-support integer sequence is a finite integral combination of the \(e_n\) and is therefore obtained from a finite product of powers of the \(a_n\).
8. If \(C\) is the mapping cone of \(q\), then\[ \pi_1(C) \cong \pi_1(\Sigma X)/\left\langle\!\left\langle q_*(\pi_1(SX))\right\rangle\!\right\rangle . \]
Write the mapping cone as \[ C=\Sigma X\cup_q CSX. \] Use open neighborhoods of the two displayed pieces whose intersection deformation retracts onto \(SX\). The cone \(CSX\) is contractible, while the other piece deformation retracts onto \(\Sigma X\). Van Kampen therefore quotients \(\pi_1(\Sigma X)\) by the normal closure of the image of \(\pi_1(SX)\) under \(q_*\).
9. The homomorphism \(\rho\) induces a surjection\[ \boxed{ \overline\rho:\pi_1(C)\twoheadrightarrow \frac{\prod_{n=1}^{\infty}\mathbb Z} {\bigoplus_{n=1}^{\infty}\mathbb Z}. } \]
Let \[ p:\prod_{n=1}^{\infty}\mathbb Z \to \left(\prod_{n=1}^{\infty}\mathbb Z\right) /\left(\bigoplus_{n=1}^{\infty}\mathbb Z\right) \] be the quotient map. By 7, \[ (p\rho)\,q_*=0. \] The target is abelian, so \(p\rho\) kills the normal closure of \(q_*(\pi_1(SX))\). By 8 it therefore factors uniquely through \(\pi_1(C)\). The factor \(\overline\rho\) is surjective because both \(\rho\) and \(p\) are surjective.
10. The group \(\pi_1(C)\) is uncountable.
The product \[ \prod_{n=1}^{\infty}\mathbb Z \] has cardinality \(2^{\aleph_0}\), whereas the direct sum \[ \bigoplus_{n=1}^{\infty}\mathbb Z \] is countable. Thus the quotient in 9 is uncountable. Since it is a quotient of \(\pi_1(C)\), the group \(\pi_1(C)\) must also be uncountable.
11. The space \(C\) is the reduced suspension of the cone \(CX\), so a reduced suspension of a contractible space need not be contractible.
Writing out the two quotient constructions shows that forming the mapping cone of \[ SX\to\Sigma X \] amounts to taking the suspension of \(CX\) and collapsing the suspension line over the cone point; this is precisely the reduced suspension \(\Sigma(CX)\). The cone \(CX\) is contractible. But 10 gives \[ \pi_1(C)\ne0, \] so \(C\cong\Sigma(CX)\) is not contractible.