Given a map \(f: X \to Y\) and a path \(h: I \to X\) from \(x_0\) to \(x_1\), show that \(f_* \beta_h = \beta_{fh} f_*\) in the diagram:
\[\begin{array}{rcl} \pi_1(X, x_1) & \xrightarrow{\beta_h} & \pi_1(X, x_0) \\ \downarrow f_* & & \downarrow f_* \\ \pi_1(Y, f(x_1)) & \xrightarrow{\beta_{fh}} & \pi_1(Y, f(x_0)) \end{array}\]
Solution
1. Let \([\gamma]\in\pi_1(X,x_1)\).Then\[ (f_*\beta_h)([\gamma]) = [f\circ(h\cdot\gamma\cdot\bar h)]. \]
By definition, \[ \beta_h([\gamma])=[h\cdot\gamma\cdot\bar h] \] as a loop based at \(x_0\), and \(f_*\) is induced by postcomposition with \(f\).
2. One has\[ f\circ(h\cdot\gamma\cdot\bar h) = (f\circ h)\cdot(f\circ\gamma)\cdot\overline{f\circ h} \]up to the standard concatenation parametrization.
Postcomposition with \(f\) preserves each of the three pieces of the concatenation, and \[ f\circ\bar h=\overline{f\circ h}. \]
3. Therefore\[ (f_*\beta_h)([\gamma]) = (\beta_{fh}f_*)([\gamma]). \]
Using <1>1–<1>2, \[ \begin{aligned} (f_*\beta_h)([\gamma]) &=[(fh)\cdot(f\gamma)\cdot\overline{fh}]\\ &=\beta_{fh}([f\gamma])\\ &=(\beta_{fh}f_*)([\gamma]). \end{aligned} \]
4. Hence\[ f_*\beta_h=\beta_{fh}f_*. \]
The element \([\gamma]\) in 1 was arbitrary.