Functoriality tells us that homeomorphic based spaces have isomorphic fundamental groups. Much more is true: \(\pi_1\) cannot distinguish homotopy equivalent spaces. There is one subtlety. A homotopy \(H \colon X \times I \to Y\) between two maps need not fix basepoints, so it does not immediately induce anything on based homotopy classes. What it does produce is a path in \(Y\) traced out by the basepoint, and the two induced maps differ exactly by the corresponding basepoint-change isomorphism.
Let \(F \colon I \times I \to Y\) be continuous. Write \(b, r, l, t\) for the restrictions of \(F\) to the bottom, right, left and top edges of the square, each traversed in the direction of increasing coordinate. Then \(b * r \simeq_p l * t\) as paths from \(F(0,0)\) to \(F(1,1)\text{.}\)
In the square \(I \times I\) the two paths from \((0,0)\) to \((1,1)\) given by bottom-then-right and left-then-top are path homotopic, by the straight-line homotopy between them; this is legitimate because \(I \times I\) is convex and the homotopy fixes the two corners. Composing that path homotopy with \(F\) gives the required path homotopy in \(Y\text{.}\)
Let \(\varphi, \psi \colon X \to Y\) be continuous maps and let \(H \colon X \times I \to Y\) be a homotopy from \(\varphi\) to \(\psi\text{.}\) Fix \(x_0 \in X\) and let \(\alpha(t) = H(x_0, t)\) be the path traced by the basepoint, running from \(\varphi(x_0)\) to \(\psi(x_0)\text{.}\) Then
Let \(f\) be a loop in \(X\) at \(x_0\) and consider
\begin{equation*}
F \colon I \times I \to Y, \qquad F(s,t) = H\bigl(f(s), t\bigr).
\end{equation*}
Its bottom edge is \(\varphi \circ f\) and its top edge is \(\psi \circ f\text{.}\) Since \(f(0) = f(1) = x_0\text{,}\) both the left and the right edge equal \(\alpha\text{.}\)Lemma 1.17 therefore gives
If \(\varphi \colon X \to Y\) is a homotopy equivalence, then \(\varphi_* \colon \pi_1(X, x_0) \to \pi_1(Y, \varphi(x_0))\) is an isomorphism for every \(x_0 \in X\text{.}\)
Let \(\psi \colon Y \to X\) be a homotopy inverse. Applying Proposition 1.18 to a homotopy \(\psi \circ \varphi \simeq \mathrm{Id}_X\) at the basepoint \(x_0\text{,}\) and to a homotopy \(\varphi \circ \psi \simeq \mathrm{Id}_Y\) at the basepoint \(\varphi(x_0)\text{,}\) produces paths \(\alpha\) in \(X\) from \(x_0\) to \(\psi\varphi(x_0)\) and \(\beta\) in \(Y\) from \(\varphi(x_0)\) to \(\varphi\psi\varphi(x_0)\) such that both triangles in
commute, the horizontal maps being isomorphisms by Exercise 1.11. The left triangle exhibits \(\psi_*\) as a surjection, since \(\hat{\alpha}\) is; the right triangle exhibits \(\psi_*\) as an injection, since \(\hat{\beta}\) is. So \(\psi_* \colon \pi_1(Y, \varphi(x_0)) \to \pi_1(X, \psi\varphi(x_0))\) is an isomorphism, and therefore so is \(\varphi_* = \psi_*^{-1} \circ \hat{\alpha}\text{.}\)
