Define what it means for \(Y\) to be a strong deformation retract of \(X\), where \(Y\subseteq X\) are topological spaces. Prove that if \(i:Y\to X\) is the inclusion map and \(y\in Y\), then the induced homomorphism \(i_*:\pi_1(Y,y)\to\pi_1(X,y)\) is an isomorphism.
Solution
1. Definition of strong deformation retract.
A subspace \(Y\subseteq X\) is a strong deformation retract of \(X\) if there is a continuous map \[ H:X\times[0,1]\longrightarrow X \] such that \[ H(x,0)=x\qquad(x\in X), \] \[ H(x,1)\in Y\qquad(x\in X), \] and \[ H(y,t)=y\qquad(y\in Y,\;t\in[0,1]). \] Thus \(H\) deforms the identity map of \(X\) to a map with image in \(Y\), while fixing every point of \(Y\) throughout the deformation.
2. The endpoint map gives a retraction \(r:X\to Y\) satisfying\[ r\circ i=\id_Y. \]
Define \[ r(x)=H(x,1). \] The endpoint condition says \(r(x)\in Y\), so \(r\) may be regarded as a continuous map \(X\to Y\). If \(y\in Y\), the strong deformation condition gives \[ r(i(y))=H(y,1)=y. \] Hence \[ r\circ i=\id_Y. \]
3. The maps \(i\circ r\) and \(\id_X\) are homotopic through maps fixing the basepoint \(y\).
Regarding \(r\) as \(X\to Y\) and \(i\) as the inclusion, the composite \(i\circ r:X\to X\) is simply \[ x\longmapsto H(x,1). \] The given deformation \(H\) satisfies \[ H(-,0)=\id_X, \qquad H(-,1)=i\circ r. \] Thus it is a homotopy from \(\id_X\) to \(i\circ r\). Because the chosen basepoint satisfies \(y\in Y\), the strong condition gives \[ H(y,t)=y \qquad(0\le t\le1). \] Hence this homotopy is basepoint-preserving.
4. On fundamental groups,\[ r_*\circ i_*=\id_{\pi_1(Y,y)} \quad\text{and}\quad i_*\circ r_*=\id_{\pi_1(X,y)}. \]
Functoriality of the induced homomorphism gives \[ r_*\circ i_*=(r\circ i)_*. \] By 2, \(r\circ i=\id_Y\), so \[ r_*\circ i_*=(\id_Y)_*=\id_{\pi_1(Y,y)}. \]
Likewise, \[ i_*\circ r_*=(i\circ r)_*. \] By 3, \(i\circ r\) is basepoint-preserving homotopic to \(\id_X\). Basepoint-preserving homotopic maps induce the same homomorphism on the fundamental group, hence \[ (i\circ r)_*=(\id_X)_*=\id_{\pi_1(X,y)}. \] This proves both identities.
5. Therefore\[ \boxed{i_*:\pi_1(Y,y)\longrightarrow\pi_1(X,y)\text{ is an isomorphism}.} \]
By 4, \(r_*\) is simultaneously a left and right inverse for \(i_*\). Therefore \(i_*\) is an isomorphism with inverse \(r_*\).