Let \(X\) be the shrinking wedge of circles in Example 1.25, the subspace of \(\mathbb{R}^2\) consisting of the circles of radius \(1/n\) and center \((1/n, 0)\) for \(n = 1, 2, \ldots\).
If \(f_n: I \to X\) is the loop based at the origin winding once around the \(n\)th circle, show that the infinite product of commutators \([f_1, f_2][f_3, f_4] \cdots\) defines a loop in \(X\) that is nontrivial in \(H_1(X)\).
If we view \(X\) as the wedge sum of the subspaces \(A\) and \(B\) consisting of the odd-numbered and even-numbered circles, respectively, use the same loop to show that the map \(H_1(X) \to H_1(A) \oplus H_1(B)\) induced by the retractions of \(X\) onto \(A\) and \(B\) is not an isomorphism.
Solution
Let \[ \gamma=[f_1,f_2][f_3,f_4]\cdots \] be the infinite concatenation, parametrized so that the \(j\)th commutator is traversed on a subinterval whose lengths tend to \(0\). Since the diameters of the circles tend to \(0\), this defines a continuous based loop in \(X\).
For \(N\ge1\), let \[ r_N:X\longrightarrow \bigvee_{i=1}^N S_i^1 \] be the retraction that is the identity on the first \(N\) circles and collapses all later circles to the wedge point.
1. The class of \(\gamma\) is nonzero in \(H_1(X;\mathbb Z)\).
Suppose instead that \([\gamma]=0\) in \(H_1(X;\mathbb Z)\). By the Hurewicz theorem in degree \(1\), \[ H_1(X;\mathbb Z)\cong \pi_1(X)_{\mathrm{ab}}, \] so the element represented by \(\gamma\) lies in the commutator subgroup of \(\pi_1(X)\). By definition of the commutator subgroup, it is therefore a finite product of commutators: \[ [\gamma]_{\pi_1}=[u_1,v_1]\cdots [u_k,v_k] \] for some \(u_i,v_i\in\pi_1(X)\).
Apply \((r_{2k+2})_*\). If \(x_i\) denotes the standard free generator of \[ \pi_1\!\left(\bigvee_{i=1}^{2k+2}S_i^1\right)\cong F(x_1,\dots,x_{2k+2}), \] then every commutator in the tail of \(\gamma\) uses circles numbered \(>2k+2\) and is collapsed. Hence \[ (r_{2k+2})_*([\gamma]_{\pi_1}) =[x_1,x_2][x_3,x_4]\cdots[x_{2k+1},x_{2k+2}]. \] On the other hand, the image of the right-hand side above is a product of at most \(k\) commutators in this free group. This contradicts Exercise 12, which says that the displayed product of \(k+1\) basis commutators cannot be expressed as a product of fewer than \(k+1\) commutators. Therefore \([\gamma]\ne0\) in \(H_1(X;\mathbb Z)\).
2. The map\[ H_1(X)\longrightarrow H_1(A)\oplus H_1(B) \]induced by the retractions onto the odd and even circles is not an isomorphism.
Let \(r_A:X\to A\) and \(r_B:X\to B\) be the two retractions. Each factor \[ [f_{2j-1},f_{2j}] \] contains one loop in \(A\) and one loop in \(B\). Under \(r_A\), the even loop is collapsed, so the commutator becomes trivial; under \(r_B\), the odd loop is collapsed, so it again becomes trivial. Thus both \((r_A)_*([\gamma])\) and \((r_B)_*([\gamma])\) vanish in first homology.
By part 1, however, \([\gamma]\ne0\) in \(H_1(X)\). Hence the displayed map has nonzero kernel and therefore is not an isomorphism.