Show that if \((X, e)\) is an H-space then \(\pi_1(X, e)\) is abelian. [Compare the usual composition \(f \cdot g\) of loops with the product \(\mu(f(t), g(t))\) coming from the H-space multiplication \(\mu\).]
Solution
1. Setup the two binary operations on the space of based loops \(\Omega(X, e)\):
1.1. Let \((X, e, \mu)\) be an H-space with multiplication \(\mu: X \times X \to X\), so \(\mu(x, e) \simeq x\) and \(\mu(e, x) \simeq x\) via homotopies fixing the basepoint \(e\).
definition of an H-space.
1.2. Let \(f, g: [0, 1] \to X\) be based loops (\(f(0) = f(1) = g(0) = g(1) = e\)).
setup.
1.3. Define standard loop concatenation \(*\) and pointwise H-space product \(\odot\):\[ (f * g)(t) = \begin{cases} f(2t) & 0 \le t \le 1/2 \\ g(2t - 1) & 1/2 \le t \le 1, \end{cases} \qquad (f \odot g)(t) = \mu(f(t), g(t)). \]
definitions of concatenation and induced pointwise multiplication.
1.4. Let \(c_e\) denote the constant loop at \(e\).Then \(f \simeq f * c_e \simeq c_e * f\) and \(g \simeq c_e * g \simeq g * c_e\) relative to endpoints.
identity element properties in the fundamental group.
2. Show that \([f * g] = [f \odot g]\) in \(\pi_1(X, e)\):
2.1. Using the homotopies \(f \simeq f * c_e\) and \(g \simeq c_e * g\), we have \(f \odot g \simeq (f * c_e) \odot (c_e * g)\).
homotopy invariance of the induced map \(\mu_*\).
2.2. For \(t \in [0, 1/2]\):\[ ((f * c_e) \odot (c_e * g))(t) = \mu((f * c_e)(t), (c_e * g)(t)) = \mu(f(2t), e) \simeq f(2t) = (f * g)(t). \]
definition of \(*\) and unit axiom \(\mu(x, e) \simeq x\).
2.3. For \(t \in [1/2, 1]\):\[ ((f * c_e) \odot (c_e * g))(t) = \mu((f * c_e)(t), (c_e * g)(t)) = \mu(e, g(2t - 1)) \simeq g(2t - 1) = (f * g)(t). \]
definition of \(*\) and unit axiom \(\mu(e, y) \simeq y\).
2.4. Gluing the two halves gives \([f \odot g] = [f * g]\).
2 and 3.
3. Show that \([g * f] = [f \odot g]\) in \(\pi_1(X, e)\):
3.1. Using the homotopies \(f \simeq c_e * f\) and \(g \simeq g * c_e\), we have \(f \odot g \simeq (c_e * f) \odot (g * c_e)\).
homotopy invariance.
3.2. For \(t \in [0, 1/2]\):\[ ((c_e * f) \odot (g * c_e))(t) = \mu(e, g(2t)) \simeq g(2t) = (g * f)(t). \]
unit axiom \(\mu(e, y) \simeq y\).
3.3. For \(t \in [1/2, 1]\):\[ ((c_e * f) \odot (g * c_e))(t) = \mu(f(2t - 1), e) \simeq f(2t - 1) = (g * f)(t). \]
unit axiom \(\mu(x, e) \simeq x\).
3.4. Gluing the two halves gives \([f \odot g] = [g * f]\).
2 and 3.
4. Conclusion: Combining 2 and 3 yields:\[ [f][g] = [f * g] = [f \odot g] = [g * f] = [g][f]. \]Thus \(\pi_1(X, e)\) is an abelian group.
2 and 3.
Q.E.D.