Let \(A\) be a subspace of \(X\); let \(j: A \to X\) be the inclusion map, and let \(f: X \to A\) be a continuous map. Suppose there is a homotopy \(H: X \times I \to X\) between the map \(j \circ f\) and the identity map of \(X\).
Show that if \(f\) is a retraction, then \(j_*\) is an isomorphism.
Show that if \(H\) maps \(A \times I\) into \(A\), then \(j_*\) is an isomorphism.
Give an example in which \(j_*\) is not an isomorphism.
Solution
Let \(a_0\in A\), and use it as basepoint; put \(x_0=j(a_0)\). The homotopy \(H\) need not preserve the basepoint, so homotopic maps induce homomorphisms differing by a change-of-basepoint isomorphism. In particular, since \(jf\simeq\operatorname{id}_X\), the composite \[ j_*f_*:\pi_1(X,x_0)\to\pi_1(X,x_0) \] is an automorphism (up to the canonical basepoint-change isomorphism).
If \(f\) is a retraction, then \(fj=\operatorname{id}_A\), so \[ f_*j_*=\operatorname{id}_{\pi_1(A,a_0)}. \] Hence \(j_*\) is injective. Since \(j_*f_*\) is an automorphism, \(j_*\) is surjective. Thus \(j_*\) is an isomorphism.
If \(H(A\times I)\subset A\), then restricting \(H\) to \(A\times I\) gives a homotopy in \(A\) from \(fj:A\to A\) to \(\operatorname{id}_A\). Hence \((fj)_*=f_*j_*\) is an automorphism (again with the harmless basepoint change determined by the restricted homotopy). Thus \(j_*\) has both a left inverse up to automorphism and, from \(jf\simeq\operatorname{id}_X\), a right inverse up to automorphism. Consequently \(j_*\) is bijective, hence an isomorphism.
Let \(X=D^2\), \(A=S^1=\partial D^2\), and let \(f:D^2\to S^1\) be a constant map. Then \(jf\) is constant and is homotopic to \(\operatorname{id}_{D^2}\) because \(D^2\) is contractible. But \[ j_*:\pi_1(S^1)\cong\mathbb Z\longrightarrow\pi_1(D^2)=0 \] is not an isomorphism.