Proposition

The map $\beta_h: \pi_1(X, x_1) \rightarrow \pi_1(X, x_0)$ is an isomorphism.

Solution

Let $h$ be a path from $x_0$ to $x_1$. Define $\beta_h: \pi_1(X, x_1) \rightarrow \pi_1(X, x_0)$ such that $\beta_h([f]) = [h \cdot f \cdot \overline{h}]$.

• $\beta_h$ is well-defined because $h \cdot f \cdot \overline{h}$ is a loop based at $x_0$ for any $[f] \in \pi_1(X, x_1)$.

• $\beta_h$ is a homomorphism because

  \[\begin{align*} \beta_h([f] \cdot [g]) &= \beta_h([f \cdot g]) \\ &= [h \cdot f \cdot g \cdot \overline{h}] \\ &= [h \cdot f \cdot \overline{h} \cdot h \cdot g \cdot \overline{h}] \\ &= [h \cdot f \cdot \overline{h}] \cdot [h \cdot g \cdot \overline{h}] \\ &= \beta_h([f]) \cdot \beta_h([g]). \end{align*}\]
• $\beta_h(\beta_{\overline{h}}([f])) = [h \cdot \overline{h} \cdot f \cdot h \overline{h}] = [f]$.
• $\beta_{\overline{h}}(\beta_{h}([f])) = [\overline{h} \cdot h \cdot f \cdot \overline{h} \cdot h] = [f]$.

Therefore, $\beta_h$ is an isomorphism.