Given a map $f: X \rightarrow Y$ and a path $h: I \rightarrow X$ from $x_0$ to $x_1$, show that $f_*\beta_h = \beta_{fh}f_*$.
by Hidenori
Proposition
Given a map $f: X \rightarrow Y$ and a path $h: I \rightarrow X$ from $x_0$ to $x_1$, show that \(f_*\beta_h = \beta_{fh}f_*\).
Solution
Let $[g] \in \pi_1(X, x_1)$ be given. Then $g$ is a loop in $X$ based at $x_1$.
\[\begin{align*} f_*\beta_h([g]) &= f_*([h \cdot g \cdot \overline{h}]) \\ &= [f \circ (h \cdot g \cdot \overline{h})]. \\ \\ \beta_{fh}f_*([g]) &= \beta_{fh}([f \circ g]) \\ &= [(f \circ h) \cdot (f \circ g) \cdot \overline{f \circ h}]. \end{align*}\]We claim that $f \circ (h \cdot g \cdot \overline{h})$ and $(f \circ h) \cdot (f \circ g) \cdot \overline{f \circ h}$ are the same loop. Let $t \in [0, 1]$ be given.
\[\begin{align*} (f \circ (h \cdot g \cdot \overline{h}))(t) &= f((h \cdot g \cdot \overline{h})(t)) \\ &= \begin{cases} f(h(3t)) & (t \in [0, 1/3]) \\ f(g(3t - 1)) & (t \in [1/3, 2/3]) \\ f(\overline{h}(3t - 2)) & (t \in [2/3, 1]). \end{cases} \\ \\ ((f \circ h) \cdot (f \circ g) \cdot (f \circ h))(t) &= \begin{cases} f(h(3t)) & (t \in [0, 1/3]) \\ f(g(3t - 1)) & (t \in [1/3, 2/3]) \\ \overline{(f \circ h)}(3t - 2) & (t \in [2/3, 1]). \end{cases}. \end{align*}\]$\overline{(f \circ h)}(3t - 2) = (f \circ h)(1 - (3t - 2)) = (f \circ h)(3 - 3t) = f(h(3 - 3t)) = f(\overline{h}(3 t - 2))$.
Thus those two loops are identical, so \(f_*\beta_h = \beta_{fh}f_*\).
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