Show that in a homotopy-associative H-space whose set of path-components is a group with respect to the multiplication induced by the H-space structure, all the path-components must be homotopy equivalent. [Homotopy-associative means associative up to homotopy.]
Solution
Let \(X_0\) be the path-component of the identity \(e\), and let \(X_a\) be any other path-component. Choose \(x\in X_a\). Since the set of path-components is a group under multiplication, choose \(y\) in the inverse component \(X_{a^{-1}}\).
Left multiplication by \(x\) defines a map \[ L_x:X_0\to X_a, \qquad z\mapsto\mu(x,z), \] and left multiplication by \(y\) defines \[ L_y:X_a\to X_0. \] By homotopy associativity, \[ L_yL_x(z)=\mu(y,\mu(x,z)) \simeq \mu(\mu(y,x),z)=L_{yx}(z). \] The point \(\mu(y,x)\) lies in the identity component \(X_0\), so choose a path in \(X_0\) from \(\mu(y,x)\) to \(e\). Multiplication along this path gives a homotopy \[ L_{yx}\simeq L_e\simeq\operatorname{id}_{X_0}. \] Similarly, \[ L_xL_y\simeq L_{xy}\simeq\operatorname{id}_{X_a}. \] Therefore \(L_x:X_0\to X_a\) is a homotopy equivalence. Since \(X_a\) was arbitrary, all path-components are homotopy equivalent.