Deformation Retractions as a Tool for Fundamental Groups
A space can be difficult to study in its given form but have the same fundamental group as a simpler subspace. A deformation retraction provides a precise way to make this reduction: points move continuously into the subspace, while points already there stay fixed throughout the motion. We will prove that the inclusion of the subspace then induces an isomorphism on fundamental groups, and use explicit homotopies to apply the result.
For a space \(X\) and a basepoint \(x_0\in X\), \(\pi_1(X,x_0)\) is the group of based loops at \(x_0\), considered up to homotopy that keeps the endpoints at \(x_0\). A continuous map \(f:(X,x_0)\to(Y,y_0)\) sends a loop \(\gamma\) to \(f\circ\gamma\) and induces a homomorphism \(f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)\). We use the identity and composition laws for induced homomorphisms.
Write \(i:A\to X\) for the inclusion. A deformation retraction has \(r\circ i=\operatorname{id}_A\). Its stronger feature is that \(i\circ r\), while not necessarily equal to \(\operatorname{id}_X\), is joined to \(\operatorname{id}_X\) by a homotopy that fixes every point of \(A\).
The Induced Isomorphism
Proof. Since \(r(a_0)=a_0\), both \(i\) and \(r\) preserve the chosen basepoint and their induced homomorphisms have the stated domains and codomains. From \(r\circ i=\operatorname{id}_A\) and the composition law, we obtain \(r_*\circ i_*=(r\circ i)_*=(\operatorname{id}_A)_*=\operatorname{id}_{\pi_1(A,a_0)}\).
For the other composition, let \(\gamma:[0,1]\to X\) be any loop based at \(a_0\). Define \(G:[0,1]\times[0,1]\to X\) by \(G(s,t)=H(\gamma(s),t)\). This is continuous as a composition of continuous maps. At \(t=0\), \(G(s,0)=\gamma(s)\); at \(t=1\), \(G(s,1)=r(\gamma(s))=(i\circ r\circ\gamma)(s)\), where the last expression views the result as a loop in \(X\). For every \(t\), \(G(0,t)=H(a_0,t)=a_0\) and \(G(1,t)=H(a_0,t)=a_0\). Thus \(G\) is a homotopy of based loops, so \([i\circ r\circ\gamma]=[\gamma]\). This holds for every based loop, hence \((i\circ r)_*=\operatorname{id}_{\pi_1(X,a_0)}\). By composition, \(i_*\circ r_*=\operatorname{id}_{\pi_1(X,a_0)}\). The two identities show that \(i_*\) and \(r_*\) are inverse isomorphisms. \(\square\)
The basepoint condition matters: the theorem uses a point \(a_0\) in the retract, and the deformation homotopy fixes that point. The proof does not require \(X\) or \(A\) to be path-connected; it gives the asserted isomorphism at each chosen basepoint in \(A\).
Worked Example: An Open Strip Retracts onto a Line
Let \(X=\mathbb{R}\times(-1,1)\) with the subspace topology from \(\mathbb{R}^2\), and let \(A=\mathbb{R}\times\{0\}\), also with its subspace topology. Define \(r:X\to A\) by \(r(x,y)=(x,0)\), and set \(H((x,y),t)=(x,(1-t)y)\).
For \((x,y)\in X\) and \(t\in[0,1]\), \(0\leq 1-t\leq1\). If \(y\in(-1,1)\), then \(-1<(1-t)y<1\): when \(y\geq0\), \(0\leq(1-t)y\leq y<1\); when \(y<0\), \(y\leq(1-t)y\leq0\), so \(-1<y\leq(1-t)y\). Thus \(H\) takes values in \(X\). Its coordinate formula is continuous, \(H((x,y),0)=(x,y)\), and \(H((x,y),1)=(x,0)=r(x,y)\). For \((x,0)\in A\), \(H((x,0),t)=(x,0)\) for every \(t\). Therefore \(A\) is a deformation retract.
Check the formula. At \((2,\tfrac12)\) and \(t=\tfrac13\), the second coordinate is \((1-\tfrac13)\tfrac12=\tfrac23\cdot\tfrac12=\tfrac13\), so \(H((2,\tfrac12),\tfrac13)=(2,\tfrac13)\), which is still in \(X\). At \(t=1\), the same point maps to \((2,0)\), and every point on \(A\) remains unchanged.
The line \(A\) is convex in \(\mathbb{R}^2\), so its fundamental group is trivial by the result for convex spaces. The theorem therefore gives \(\pi_1(X,(x_0,0))\cong\pi_1(A,(x_0,0))\), which is the trivial group for every \(x_0\in\mathbb{R}\).
Radial Retractions and the Circle
For \(x=(x_1,x_2)\in\mathbb{R}^2\), write \(\|x\|=\sqrt{x_1^2+x_2^2}\), and let \(S^1=\{x\in\mathbb{R}^2:\|x\|=1\}\) with its subspace topology. A radial homotopy changes the distance from the origin while keeping the direction fixed. The homotopy formulas must be checked both to stay in the space and to fix the proposed retract.
Worked Example: The Punctured Plane Retracts onto the Unit Circle
Give \(X=\mathbb{R}^2\setminus\{(0,0)\}\) the subspace topology from \(\mathbb{R}^2\). Regard \(A=S^1\) as a subspace of \(X\). Define \(r(x)=x/\|x\|\) and \(H(x,t)=((1-t)\|x\|+t)x/\|x\|\).
For \(x\in X\), \(\|x\|>0\). The scalar \((1-t)\|x\|+t\) is positive: if \(t=0\) it equals \(\|x\|>0\); if \(t>0\), it is a sum of nonnegative terms with \(t>0\). Thus \(H(x,t)\ne(0,0)\) throughout. The formula is continuous on \(X\times[0,1]\), because division by \(\|x\|\) is continuous where \(\|x\|>0\). At \(t=0\), \(H(x,0)=x\), and at \(t=1\), \(H(x,1)=x/\|x\|=r(x)\), which has norm \(1\). If \(a\in S^1\), then \(\|a\|=1\), so \(H(a,t)=((1-t)+t)a=a\). These checks prove that \(A\) is a deformation retract.
Check with a point. For \(x=(3,4)\), \(\|x\|=\sqrt{3^2+4^2}=5\). At \(t=\tfrac12\), the new radius is \((1-\tfrac12)5+\tfrac12=\tfrac52+\tfrac12=3\), and the direction is \(x/\|x\|=(\tfrac35,\tfrac45)\). Hence \(H((3,4),\tfrac12)=(\tfrac95,\tfrac{12}5)\), whose norm is \(\sqrt{(\tfrac95)^2+(\tfrac{12}5)^2}=\sqrt{\tfrac{81+144}{25}}=3\). At \(t=1\) the formula gives \((\tfrac35,\tfrac45)\), of norm \(1\), as required.
With basepoint \((1,0)\in S^1\), the theorem yields \(\pi_1(X,(1,0))\cong\pi_1(S^1,(1,0))\). This reduces the punctured-plane computation to the circle; the deformation-retraction argument itself does not calculate the group of the circle.
Worked Example: A Cylinder Retracts onto a Circle
Give \(S^1\times\mathbb{R}\) the product topology, and let \(A=S^1\times\{0\}\) with its subspace topology. Define \(r(z,s)=(z,0)\) and \(H((z,s),t)=(z,(1-t)s)\). The coordinate formulas are continuous. For every \(s\in\mathbb{R}\), \((1-t)s\in\mathbb{R}\), so the homotopy stays in the cylinder. It starts at \((z,s)\), ends at \((z,0)\), and fixes every \((z,0)\) for all \(t\). Thus this is a deformation retraction.
Check with a point. For any \(z\in S^1\), the point \((z,3)\) maps at \(t=\tfrac23\) to \((z,(1-\tfrac23)3)=(z,1)\), and at \(t=1\) to \((z,0)\). If the input is already \((z,0)\), then its second coordinate stays \((1-t)0=0\) for every \(t\).
The map \(z\mapsto(z,0)\) is a homeomorphism from \(S^1\) to \(A\): it is continuous, bijective, and its inverse is the restriction of the first-coordinate projection, which is continuous. Homeomorphism invariance and the deformation-retraction theorem therefore give \(\pi_1(S^1\times\mathbb{R},(z_0,0))\cong\pi_1(S^1,z_0)\) for every \(z_0\in S^1\).
What the Retraction Alone Does Not Show
Common pitfall. From \(r\circ i=\operatorname{id}_A\), one gets \(r_*\circ i_*=\operatorname{id}\), but this identity alone does not show that \(i_*\) is surjective. The algebraic example \(i:\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}\), \(i(n)=(n,0)\), and \(r:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}\), \(r(m,n)=m\), makes the issue concrete: \(r\circ i\) is the identity, but \(i\) is not surjective, since \((0,1)\) is not in its image. Therefore the mistaken inference “\(r_*\circ i_*=\operatorname{id}\), so \(i_*\) is an isomorphism” is invalid. For a deformation retract, the additional based homotopy proves \(i_*\circ r_*=\operatorname{id}\) as well; that second identity supplies surjectivity.
In practice, the proof has two complementary checks: verify that the retraction fixes the subspace, and verify that the homotopy fixes it at every time. A homotopy that only ends in \(A\), but moves points of \(A\) along the way, does not meet the definition used here, and the based-loop argument above would not follow for a basepoint in \(A\).
Check Your Understanding
For each question, identify the basepoint and check the relevant homotopy conditions explicitly.
- In the open strip example, verify directly that \(r\) is continuous and that \(r|_A=\operatorname{id}_A\).
- Explain why the radial formula for the punctured plane cannot be evaluated at the origin.
- For a deformation retract \(A\subseteq X\), which composition of induced homomorphisms follows directly from \(r\circ i=\operatorname{id}_A\), and which uses the deformation homotopy?
- Write a deformation retraction of \(\mathbb{R}^2\) onto the \(x\)-axis and check that it fixes the axis throughout.
- Why does the theorem state its isomorphism at a basepoint belonging to \(A\)?