Prove the following.
Theorem. Let \(X\) be a space that is path connected, locally path connected, and semilocally simply connected. If \(X\) is regular with a countable basis, then \(\pi_1(X, x_0)\) is countable.
Let \(\mathcal{A}\) be a covering of \(X\) by path-connected open sets such that for each \(A \in \mathcal{A}\) and each \(a \in A\), the homomorphism \(\pi_1(A, a) \to \pi_1(X, a)\) induced by inclusion is trivial. Let \(\mathcal{B}\) be a countable open covering of \(X\) by nonempty path-connected sets that satisfies the conditions of E-8Q9TN. Choose a point \(p(B) \in B\) for each \(B \in \mathcal{B}\). For each pair \(B\), \(B'\) of elements of \(\mathcal{B}\) for which \(B \cap B' \neq \varnothing\), choose a path \(g(B, B')\) in \(B \cup B'\) from \(p(B)\) to \(p(B')\). We call the path \(g(B, B')\) a select path.
Let \(B_0\) be a fixed element of \(\mathcal{B}\); let \(x_0 = p(B_0)\). Show that if \(f\) is a loop in \(X\) based at \(x_0\), then \(f\) is path homotopic to a product of select paths, as follows:
- Show that there is a subdivision
\[ 0 = t_0 < \dots < t_n = 1 \]
of \([0, 1]\) such that \(f\) maps \([t_{n-1}, t_n]\) into \(B_0\), and for each \(i = 1, \ldots, n-1\), \(f\) maps \([t_{i-1}, t_i]\) into an element \(B_i\) of \(\mathcal{B}\). Set \(B_n = B_0\).
- Let \(f_i\) be the positive linear map of \([0, 1]\) onto \([t_{i-1}, t_i]\) followed by \(f\). Let \(g_i = g(B_{i-1}, B_i)\). Choose a path \(\alpha_i\) in \(B_i\) from \(f(t_i)\) to \(p(B_i)\); if \(i = 0\) or \(n\), let \(\alpha_i\) be the constant path at \(x_0\). Show that
\[ [f_i] * [\alpha_i] = [\alpha_{i-1}] * [g_i]. \]
- Show that \([f] = [g_1] * \dots * [g_n]\).
Solution
Let \(\mathcal A\) and the countable star-fine path-connected cover \(\mathcal B\) be as in the statement. Thus whenever \(B,B'\in\mathcal B\) intersect, there is \(A\in\mathcal A\) with \[ B\cup B'\subset A, \] and every loop in \(A\) maps trivially in \(\pi_1(X)\).
The sets \(f^{-1}(B)\), \(B\in\mathcal B\), form an open cover of the compact interval \([0,1]\). By the Lebesgue-number/subdivision lemma, there is a subdivision \[ 0=t_0<\cdots<t_n=1 \] such that each \(f([t_{i-1},t_i])\) lies in some \(B_i\in\mathcal B\). Since \(f(1)=x_0\in B_0\) and \(B_0\) is open, refine the subdivision near \(1\) so that the last interval maps into \(B_0\); set \(B_n=B_0\). Consecutive \(B_{i-1}\) and \(B_i\) intersect because both contain \(f(t_{i-1})\) after the evident index shift (equivalently choose each \(B_i\) around the common subdivision point during refinement).
The paths \[ f_i*\alpha_i \qquad\text{and}\qquad \alpha_{i-1}*g_i \] have the same initial point \(f(t_{i-1})\) and terminal point \(p(B_i)\). Their concatenation with one reversed is a loop contained in \[ B_{i-1}\cup B_i. \] Because \(B_{i-1}\cap B_i\ne\varnothing\), this union lies in some \(A\in\mathcal A\). The inclusion-induced map \(\pi_1(A)\to\pi_1(X)\) is trivial, so the loop is nullhomotopic in \(X\). Hence the two paths are homotopic rel endpoints: \[ [f_i]*[\alpha_i]=[\alpha_{i-1}]*[g_i]. \]
Rearranging the equality from (b), \[ [f_i]=[\alpha_{i-1}]*[g_i]*[\alpha_i]^{-1}. \] Multiplying these identities for \(i=1,\dots,n\), all intermediate \(\alpha_i\)’s cancel. Since \(\alpha_0\) and \(\alpha_n\) are constant at \(x_0\), \[ [f]=[f_1]*\cdots*[f_n] =[g_1]*\cdots*[g_n]. \] There are only countably many select paths because \(\mathcal B\) is countable, and the set of all finite words in a countable alphabet is countable. Every element of \(\pi_1(X,x_0)\) is represented by such a finite product. Therefore \[ \boxed{\pi_1(X,x_0)\text{ is countable}.} \]