Show that the change-of-basepoint homomorphism \(\beta_h\) depends only on the homotopy class of \(h\).
Solution
Let \(h,k:I\to X\) be paths from \(x_0\) to \(x_1\) that are homotopic relative to their endpoints. Thus there is a continuous map \[ H:I\times I\to X \] such that \[ H(s,0)=h(s),\qquad H(s,1)=k(s),\qquad H(0,t)=x_0,\qquad H(1,t)=x_1. \]
1. Fix a loop \(f:I\to X\) based at \(x_1\).For each \(t\in I\), define a loop \(F_t\) based at \(x_0\) by\[ F_t(s)= \begin{cases} H(3s,t),&0\le s\le \frac13,\\ f(3s-1),&\frac13\le s\le \frac23,\\ H(3-3s,t),&\frac23\le s\le1. \end{cases} \]
At \(s=1/3\), the first and second formulas both equal \(x_1\), since \[ H(1,t)=x_1=f(0). \] At \(s=2/3\), the second and third formulas both equal \(x_1\), since \[ f(1)=x_1=H(1,t). \] Hence the three formulas paste to a continuous path. Moreover, \[ F_t(0)=H(0,t)=x_0=H(0,t)=F_t(1), \] so \(F_t\) is a loop based at \(x_0\) for every \(t\).
2. The map\[ F:I\times I\to X,\qquad F(s,t)=F_t(s), \]is a homotopy of loops relative to the basepoint from\[ h\cdot f\cdot\bar h \quad\text{to}\quad k\cdot f\cdot\bar k. \]
On each of the three closed vertical strips in \(I\times I\), the formula in 1 is continuous, and the formulas agree on the common boundaries; the pasting lemma gives continuity of \(F\). At \(t=0\), the first and third pieces are \(h\) and \(\bar h\), while at \(t=1\) they are \(k\) and \(\bar k\). The endpoint computation in 1 shows that the basepoint \(x_0\) is fixed throughout the homotopy.
3. For every \([f]\in\pi_1(X,x_1)\),\[ \beta_h([f])=\beta_k([f]). \]
By definition, \[ \beta_h([f])=[h\cdot f\cdot\bar h], \qquad \beta_k([f])=[k\cdot f\cdot\bar k]. \] These based loops are homotopic relative to the basepoint by 2, so they determine the same element of \(\pi_1(X,x_0)\).
4. Therefore \(\beta_h=\beta_k\), so the change-of-basepoint homomorphism depends only on the homotopy class of \(h\).
The equality in 3 holds for every element of \(\pi_1(X,x_1)\).