Proposition

Suppose that $X$ is a topological space, and $g$ is any path in $X$ from $p$ to $q$. Let $\Phi_g: \pi_1(X, p) \rightarrow \pi_1(X, q)$ denote the group isomorphism induced by $g$.

  1. Show that if $h$ is another path in $X$ starting at $q$, then $\Phi_{g \cdot h} = \Phi_h \circ \Phi_g$.
  2. Suppose that $\psi: X \rightarrow Y$ is continuous, and show that the following diagram commutes:

$\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x27f6}$

\[\begin{gather*} \pi_1(X, p) & \xRightarrow{\psi_*} & \pi_1(Y, \psi(p)) \\ \Phi_g \Big\downarrow & & \Big\downarrow\Phi_{\psi \circ g} \\ \pi_1(X, q) & \xRightarrow{\psi_*} & \pi_1(Y, \psi(q)). \end{gather*}\]

Solution

1

Let $h$ be a path starting at $q$. We claim that $\overline{g \cdot h} = \overline{h} \cdot \overline{g}$.

When $t \in [0, 1/2]$,

\[\begin{align*} \overline{g \cdot h}(t) &= (g \cdot h)(1 - t) \\ &= h(2(1 - t) - 1) \\ &= h(1 - 2t) \\ &= \overline{h}(2t) \\ &= (\overline{h} \cdot \overline{g})(t). \end{align*}\]

Similarly, when $t \in [1/2, 1]$, $\overline{g \cdot h}(t) = (\overline{h} \cdot \overline{g})(t)$.

For any $[f] \in \pi_1(X, p)$,

\[\begin{align*} \Phi_{g \cdot h}([f]) &= [\overline{g \cdot h}] \cdot [f] \cdot [g \cdot h] \\ &= [\overline{h} \cdot \overline{g}] \cdot [f] \cdot [g \cdot h] \\ &= [\overline{h}] \cdot [\overline{g}] \cdot [f] \cdot [g] \cdot [h] \\ &= [\overline{h}] \cdot \Phi_g([f]) \cdot [h] \\ &= \Phi_h(\Phi_g([f])) \\ &= (\Phi_h \circ \Phi_g)([f]). \end{align*}\]

Therefore, $\Phi_{g \cdot h} = \Phi_h \circ \Phi_g$.

2

Let $[f] \in \pi_1(X, p)$ be chosen arbitrarily.

\[\begin{align*} \psi_*(\Phi_g([f])) &= \psi_*([\overline{g} \cdot f \cdot g]) \\ &= [\psi(\overline{g} \cdot f \cdot g)] \\ &= [(\psi \circ \overline{g}) \cdot (\psi \circ f) \cdot (\psi \circ g)] & \text{(as shown below)} \\ &= [\psi \circ \overline{g}] \cdot [\psi \circ f] \cdot [\psi \circ g] \\ &= [\overline{\psi \circ g}] \cdot [\psi \circ f] \cdot [\psi \circ g] & \text{(as shown below)} \\ &= \Phi_{\psi \circ g}([\psi \circ f]) \\ &= \Phi_{\psi \circ g}(\psi_*([f])) \end{align*}\]

since, for any $t \in [0, 1]$,

\[\begin{align*} (\psi \circ (\overline{g} \cdot f \cdot g))(t) &= \psi((\overline{g} \cdot f \cdot g)(t)) \\ &= \begin{cases} \psi(\overline{g}(t)) & (t \in [0, 1/3]) \\ \psi(f(t)) & (t \in [1/3, 2/3]) \\ \psi(g(t)) & (t \in [2/3, 1]) \end{cases} \\ &= \psi(\overline{g}(t)) \cdot \psi(f(t)) \cdot \psi(g(t)), \end{align*}\]

and

$\overline{\psi \circ g}(t) = \psi(g(1 - t)) = \psi(\overline{g}(t)) = (\psi \circ \overline{g})(t)$.