Show the sequence \(\pi_1(X, x_0) \to \pi_1(X, A, x_0) \overset{\partial}{\longrightarrow} \pi_0(A, x_0) \to \pi_0(X, x_0)\) is exact.
Solution
For \(\pi_1(X,A,x_0)\), use relative paths \(u:I\to X\) with \(u(0)=x_0\) and \(u(1)\in A\). The boundary map is \[ \partial[u]=[u(1)]\in\pi_0(A,x_0). \]
At \(\pi_1(X,A,x_0)\): a relative class \([u]\) lies in \(\ker\partial\) exactly when \(u(1)\) is in the component of \(x_0\) in \(A\). Choose a path \(v\subset A\) from \(u(1)\) to \(x_0\). Then the loop \(uv\) maps to the same relative class as \(u\). Hence \[ \ker\partial=\operatorname{im}\bigl(\pi_1(X,x_0)\to\pi_1(X,A,x_0)\bigr). \]
At \(\pi_0(A,x_0)\): a component \(C\) of \(A\) lies in the image of \(\partial\) exactly when there is a path in \(X\) from \(x_0\) to a point of \(C\), which is exactly the condition that \(C\) maps to the distinguished component of \(X\). Hence \[ \operatorname{im}\partial =\ker\bigl(\pi_0(A,x_0)\to\pi_0(X,x_0)\bigr). \] Thus the displayed sequence is exact as a sequence of pointed sets.