Show that there are no retractions \(r: X \to A\) in the following cases:
\(X = \mathbb{R}^3\) with \(A\) any subspace homeomorphic to \(S^1\).
\(X = S^1 \times D^2\) with \(A\) its boundary torus \(S^1 \times S^1\).
\(X = S^1 \times D^2\) and \(A\) the circle shown in the figure.
\(X = D^2 \lor D^2\) with \(A\) its boundary \(S^1 \lor S^1\).
\(X\) a disk with two points on its boundary identified and \(A\) its boundary \(S^1 \lor S^1\).
\(X\) the Möbius band and \(A\) its boundary circle.
Solution
We use the same elementary obstruction in all six cases.
1. If \(r:X\to A\) is a retraction and \(i:A\hookrightarrow X\) is the inclusion, then\[ r_*\circ i_*=\operatorname{id}_{\pi_1(A)}. \]In particular, \(i_*\) must be injective.
A retraction satisfies \[ r\circ i=\operatorname{id}_A. \] Functoriality of the fundamental group gives \[ (r\circ i)_*=r_*\circ i_*=(\operatorname{id}_A)_*. \] Any homomorphism possessing a left inverse is injective.
2. In case (a), no retraction \(\mathbb R^3\to A\cong S^1\) exists.
The space \(\mathbb R^3\) is contractible, so \[ \pi_1(\mathbb R^3)=0, \] whereas \[ \pi_1(A)\cong\mathbb Z. \] Thus the inclusion-induced map \[ i_*:\mathbb Z\to0 \] is not injective, contradicting 1 if a retraction existed.
3. In case (b), no retraction from the solid torus \(S^1\times D^2\) to its boundary torus exists.
Let \(m\subset S^1\times S^1\) be a meridian circle \[ m=\{x_0\}\times S^1. \] It represents a nontrivial element of \[ \pi_1(S^1\times S^1)\cong\mathbb Z^2, \] but in the solid torus it bounds the disk \[ \{x_0\}\times D^2. \] Hence \(i_*([m])=1\), so \(i_*\) is not injective. By 1 there can be no retraction.
4. In case (c), the indicated interior circle \(A\) is a meridian and hence no retraction \(S^1\times D^2\to A\) exists.
The dark circle in the source figure is an interior meridian of the solid torus. It bounds a meridional disk lying in \(S^1\times D^2\). Thus its generator \[ [A]\in\pi_1(A)\cong\mathbb Z \] maps to the identity in \(\pi_1(S^1\times D^2)\). The inclusion-induced map is therefore not injective, so 1 rules out a retraction.
5. In case (d), no retraction\[ D^2\vee D^2\to S^1\vee S^1 \]exists.
The wedge \(D^2\vee D^2\) is contractible: contract each disk to the common wedge point. Hence \[ \pi_1(D^2\vee D^2)=0. \] On the other hand, \[ \pi_1(S^1\vee S^1)\cong F(a,b) \] is the free group on two generators and is nontrivial. Thus the inclusion-induced map is not injective, contradicting 1.
6. In case (e), no retraction from the quotient disk to its boundary \(S^1\vee S^1\) exists.
Let the two identified boundary points divide the original boundary circle into arcs whose images become the two loops \(a\) and \(b\) of \[ A=S^1\vee S^1. \] Then \[ \pi_1(A)\cong F(a,b), \] and the element \(ab\) is nontrivial in this free group.
But the loop \(ab\) is exactly the image of the original boundary circle of \(D^2\). It bounds the image of the disk itself in the quotient space \(X\), so \[ i_*([ab])=1\in\pi_1(X). \] Hence \(i_*\) has nontrivial kernel, and 1 excludes a retraction.
7. In case (f), no retraction from the Möbius band \(M\) to its boundary circle exists.
The Möbius band deformation retracts onto its core circle, so \[ \pi_1(M)\cong\mathbb Z. \] Its boundary circle traverses the core twice. Thus, after choosing generators, the inclusion induces \[ i_*:\mathbb Z\to\mathbb Z, \qquad i_*(n)=2n. \] If a retraction existed, 1 would give a homomorphism \[ r_*:\mathbb Z\to\mathbb Z \] with \[ r_*\circ i_*=\operatorname{id}_{\mathbb Z}. \] Every homomorphism \(\mathbb Z\to\mathbb Z\) is multiplication by some integer \(k\), so this would require \[ 2k=1, \] which is impossible.
8. Therefore none of the six indicated inclusions admits a retraction.
Cases (a)–(f) are established in <1>2–<1>7.