Proposition

Let $\phi_t: X \rightarrow Y$ is a homotopy and $h$ is the path $\phi_t(x_0)$ formed by the images of a basepoint $x_0 \in X$. Then \(\phi_{0*} = \beta_h \phi_{1*}\).

Solution

Let $[f] \in \pi_1(X, x_0)$. We claim that \(\phi_{0*}([f]) = \beta_h(\phi_{1*}([f])\).

Let $F(x, t) = \phi_t(x)$ be the associated map of $\phi_t$’s. Then $F$ is continuous since $\phi_t$ is a homotopy.

$h$ is defined to be $h(t) = \phi_t(x_0)$. Define $H(s, t) = h(ts)$. Then $H$ maps $I \times I \rightarrow Y$. Then $H$ is a composition of $F$ and $(x, t) \mapsto (x_0, t)$. The latter is a continuous function because each coordinate function is continuous, so $H$ is continuous.

Consider $G: I \times I \rightarrow Y$ such that

This is well-defined because:

• $H(3(1/3), t) = H(1, t) = h(t) = \phi_t(x_0) = \phi_t(f(0)) = F(f(0), t) = F(f(3(1/3) - 1), t)$.
• $F(f(3(2/3) - 1), t) = F(f(1), t) = F(x_0, t) = \phi_t(x_0) = h(t) = H(1, t) = H(3 - 3(2/3), t)$.

Moreover, $G$ is continuous in each $[0, 1/3], [1/3, 2/3]$, and $[2/3, 1]$ since each $H, F, f$ are continuous. By the pasting lemma, $G$ is continuous.

We claim that $g_t(s) = G(s, t)$ is a homotopy from $\phi_0 \circ f$ to $h \cdot (\phi_1 \circ f) \cdot \overline{h}$.

• $g_0(s) = G(s, 0)$ is homotopic to $\phi_0 \circ f$.
• $g_1(s) = G(s, 1)$ is homotopic to $h \cdot (\phi_1 \circ f) \cdot \overline{h}$.
• $g_t(0) = G(0, t) = H(0, t) = h(0) = \phi_0(x_0)$.
• $g_t(1) = G(1, t) = H(3 - 3, t) = h(0) = \phi_0(x_0)$.

Therefore, $G$ is indeed a homotopy, so \([\phi_0 \circ f] = [h \cdot (\phi_1 \circ f) \cdot \overline{h}] = \beta_{h*}([\phi_1 \circ f])\). Thus \(\phi_{0*}([f]) = \beta_{h*}(\phi_{1*}([f])) = (\beta_{h*} \circ \phi_{1*})([f])\).