Show that if \(X\) is path connected, the homomorphism induced by a continuous map is independent of base point, up to isomorphisms of the groups involved. More precisely, let \(h: X \to Y\) be continuous, with \(h(x_0) = y_0\) and \(h(x_1) = y_1\). Let \(\alpha\) be a path in \(X\) from \(x_0\) to \(x_1\), and let \(\beta = h \circ \alpha\). Show that
\[ \hat{\beta} \circ (h_{x_0})_* = (h_{x_1})_* \circ \hat{\alpha}. \]
This equation expresses the fact that the following diagram of maps “commutes.”
\[ \begin{array}{c} \pi_1(X, x_0) \xrightarrow{\ (h_{x_0})_*\ } \pi_1(Y, y_0) \\ \Biggl\downarrow \hat{\alpha} \qquad\qquad\qquad\quad \Biggl\downarrow \hat{\beta} \\ \pi_1(X, x_1) \xrightarrow{\ (h_{x_1})_*\ } \pi_1(Y, y_1) \end{array} \]
Solution
Let \([f]\in\pi_1(X,x_0)\). By definition, \[ \widehat\alpha([f])=[\bar\alpha*f*\alpha]. \] Applying \((h_{x_1})_*\) gives \[ (h_{x_1})_*\widehat\alpha([f]) =[h\circ\bar\alpha*(h\circ f)*(h\circ\alpha)]. \] Since \(\beta=h\circ\alpha\), we have \(h\circ\bar\alpha=\bar\beta\). Hence \[ (h_{x_1})_*\widehat\alpha([f]) =[\bar\beta*(h\circ f)*\beta]. \] On the other hand, \[ \widehat\beta (h_{x_0})_*([f]) =\widehat\beta([h\circ f]) =[\bar\beta*(h\circ f)*\beta]. \] Thus the two maps agree on every class: \[ \boxed{\widehat\beta\circ(h_{x_0})_*=(h_{x_1})_*\circ\widehat\alpha}. \] Since change-of-basepoint maps are isomorphisms, this proves that the homomorphism induced by \(h\) is independent of the chosen basepoint up to the canonical basepoint-change isomorphisms.