(May 2017) If \(H:X\times[0,1]\to X\) is a homotopy with \(H_0=H_1\) the identity map, show that the map \(\gamma:I\to X\) given by \(\gamma(t)=H(x_0,t)\) is a loop representing an element \(g=[\gamma]\in\pi_1(X,x_0)\) which lies in the center of \(\pi_1(X,x_0)\), i.e. \(gh=hg\) for all \(h\in\pi_1(X,x_0)\).
Solution
1. \(\gamma\) is a loop based at \(x_0\).
\(\gamma(0) = H_0(x_0) = x_0\) and \(\gamma(1) = H_1(x_0) = x_0\) because \(H_0 = H_1 = \operatorname{id}_X\).
2. For every loop \(\alpha\) at \(x_0\), the map \(F\colon I \times I \to X\), \(F(s, t) = H(\alpha(s), t)\), has bottom and top edges \(\alpha\) and left and right edges \(\gamma\).
\(F(s, 0) = H_0(\alpha(s)) = \alpha(s)\) and \(F(s, 1) = H_1(\alpha(s)) = \alpha(s)\). Since \(\alpha(0) = \alpha(1) = x_0\), \(F(0, t) = F(1, t) = H(x_0, t) = \gamma(t)\).
3. For every loop \(\alpha\) at \(x_0\), \(\alpha \cdot \gamma \simeq \gamma \cdot \alpha\) relative to the endpoints.
For a map \(F\colon I \times I \to X\), the path along the bottom edge followed by the right edge and the path along the left edge followed by the top edge are paths in the convex square from \((0,0)\) to \((1,1)\), hence homotopic in \(I \times I\) relative to the endpoints; composing with \(F\) gives a homotopy relative to the endpoints between their images. By step 2 these images are \(\alpha \cdot \gamma\) and \(\gamma \cdot \alpha\).
4. Q.E.D.
By step 3, \([\alpha][\gamma] = [\gamma][\alpha]\) for every \(h = [\alpha] \in \pi_1(X, x_0)\), so \(g = [\gamma]\) lies in the center of \(\pi_1(X, x_0)\).