Proposition

If $\phi: X \rightarrow Y$ is a homotopy equivalence, then the induced homomorphism \(\phi_{*}: \pi_1(X, x_0) \rightarrow \pi_1(Y, \phi(x_0))\) is an isomorphism for all $x_0 \in x$.

Solution

Since $\phi: X \rightarrow Y$ is a homotopy equivalence, the following functions must exist:

• $\psi: Y \rightarrow X$, and $\psi$ is continuous.
• $\alpha_t: X \rightarrow X$.
  • The associated map $(x, t) \rightarrow \alpha_t(x)$ is continuous.
  • $\alpha_0 = \psi \circ \phi$.
  • $\alpha_1 = \mathbb{1}$.
• $\beta_t: Y \rightarrow Y$.
  • The associated map $(y, t) \rightarrow \beta_t(x)$ is continuous.
  • $\beta_0 = \phi \circ \psi$.
  • $\beta_1 = \mathbb{1}$.

Let $x_0 \in X$ be given. We will show that \(\phi_*: \pi_1(X, x_0) \rightarrow \pi_1(Y, \phi(x_0))\) is an isomorphism.

\(\phi_*\) is induced by a continuous map, so it is a homomorphism. Similarly, \(\psi_*: \pi_1(Y, \phi(x_0)) \rightarrow \pi_1(X, \psi(\phi(x_0)))\) is a homomorphism.

Consider \(\psi_* \circ \phi_*\). For any $[f] \in \pi_1(X, x_0)$, \((\psi_* \circ \phi_*)([f]) = [(\psi \cdot \phi)(f)] = [\alpha_0(f)] = \alpha_{0*}([f])\). By this lemma, \(\alpha_{0*} = \beta_h \alpha_{1*}\) where $h$ is the path such that $h(t) = \alpha_t(x_0)$. Since $\alpha_{1}$ is the identity function on $X$, $\alpha_{1*}$ is the identity function on $\pi_1(X, x_0)$. Therefore, \(\alpha_{0*} = \beta_h\). We have proved that a change-of-basepoint map is an isomorphism., so \(\alpha_{0*}\) is an isomorphism. Thus \(\psi_* \circ \phi_*\) is an isomorphism. This implies that \(\phi_*: \pi_1(X, x_0) \rightarrow \pi_1(Y, \phi(x_0))\) is injective.

Using the same argument with \(\psi_*: \pi_1(Y, \phi(x_0)) \rightarrow \pi_1(X, \psi(\phi(x_0)))\) and \(\phi_*: \pi_1(X, \psi(\phi(x_0))) \rightarrow \pi_1(Y, \phi(\psi(\phi(x_0))))\), \(\phi_* \circ \psi_*\) is an isomorphism, so \(\psi_*: \pi_1(Y, \phi(x_0)) \rightarrow \pi_1(Y, \psi(\phi(x_0)))\) is injective.

Lastly, we will show that \(\phi_*: \pi_1(X, x_0) \rightarrow \pi_1(Y, \phi(x_0))\) is surjective. Let $[f] \in \pi_1(Y, \phi(x_0))$. Then \(\psi_*([f]) = [\psi \circ f] \in \pi_1(Y, \psi(\phi(x_0)))\). Since \(\psi_* \circ \phi_*\) is an isomorphism, there exists \((\psi_* \circ \phi_*)^{-1}(\psi_*([f])) \in \pi_1(X, x_0)\). Thus $\pi_1(X, x_0)$ contains \(\phi_*^{-1}([f])\), so \(\phi_*\) is surjective.

Therefore, \(\phi_*\) is an isomorphism.