Let \(A \subset X\); suppose \(r: X \to A\) is a continuous map such that \(r(a) = a\) for each \(a \in A\). (The map \(r\) is called a retraction of \(X\) onto \(A\).) If \(a_0 \in A\), show that
\[ r_*: \pi_1(X, a_0) \to \pi_1(A, a_0) \]
is surjective.
Solution
1. The inclusion \(i: A \hookrightarrow X\) satisfies \(r \circ i = \operatorname{id}_A\):Proof:
1.1. Let \(i: (A, a_0) \hookrightarrow (X, a_0)\) denote the canonical inclusion map.
1.2. By definition of a retraction, \(r(a) = a\) for all \(a \in A\), which means that the composition satisfies:\[r \circ i = \operatorname{id}_A: A \to A.\]
1.3. Since \(r(a_0) = a_0\) and \(i(a_0) = a_0\), both \(i\) and \(r\) are basepoint-preserving continuous maps.
2. Functoriality of the fundamental group:Proof:
2.1. The fundamental group operator \(\pi_1\) is a functor from pointed topological spaces to groups.
2.2. Applying functoriality to the identity \(r \circ i = \operatorname{id}_A\) yields:\[r_* \circ i_* = (r \circ i)_* = (\operatorname{id}_A)_* = \operatorname{id}_{\pi_1(A, a_0)},\]where \(i_*: \pi_1(A, a_0) \to \pi_1(X, a_0)\) and \(r_*: \pi_1(X, a_0) \to \pi_1(A, a_0)\).
3. Surjectivity of \(r_*\):Proof:
3.1. Let \(\gamma \in \pi_1(A, a_0)\) be an arbitrary element.
3.2. Define \(\beta = i_*(\gamma) \in \pi_1(X, a_0)\).
3.3. Applying \(r_*\) to \(\beta\):\[r_*(\beta) = r_*(i_*(\gamma)) = (r_* \circ i_*)(\gamma) = \operatorname{id}_{\pi_1(A, a_0)}(\gamma) = \gamma.\]
3.4. Thus, every element of \(\pi_1(A, a_0)\) is in the image of \(r_*\).
4. Q.E.D.Step 3 shows that \(r_*\) is surjective. Step 2 also shows that \(i_*\) is injective.