The No-Retraction Theorem for the Disk

Why the Boundary Cannot Be Retracted from the Disk

Let \(D^2=\{z\in\mathbb{C}:|z|\leq 1\}\) and \(S^1=\{z\in\mathbb{C}:|z|=1\}\), each with the subspace topology inherited from the standard topology on \(\mathbb{C}\). The boundary circle sits inside the disk, so there is an inclusion \(i:S^1\to D^2\). A retraction would be a continuous map \(r:D^2\to S^1\) that returns each boundary point to itself: \(r(i(z))=z\) for every \(z\in S^1\). We will show that this condition is incompatible with what these two spaces do to based loops.

Use \(1\in S^1\) as the basepoint. A loop based at \(1\) in a space \(X\) is a continuous map \(\alpha:[0,1]\to X\) with \(\alpha(0)=\alpha(1)=1\). Two such loops represent the same element of \(\pi_1(X,1)\) if they are homotopic through loops that keep both endpoints at \(1\). The fundamental group \(\pi_1(X,1)\) consists of these homotopy classes. The circle’s fundamental group is \(\mathbb{Z}\), with the class of \(\alpha(s)=e^{2\pi i s}\) corresponding to \(1\): its lift under \(u\mapsto e^{2\pi i u}\), starting at \(0\), ends at \(1\). In particular, this loop is not homotopic to the constant loop.

For the inclusion \(i:S^1\to D^2\), the basepoint \(1\) maps to \(1\), so it induces \(i_*:\pi_1(S^1,1)\to\pi_1(D^2,1)\). If a retraction \(r:D^2\to S^1\) existed, it would also preserve the chosen basepoint, because \(r(1)=1\). Thus \(r_*\) would map \(\pi_1(D^2,1)\) to \(\pi_1(S^1,1)\). The key facts are that every loop in the disk is null-homotopic, while the loop \(\alpha(s)=e^{2\pi i s}\) is not null-homotopic in the circle.

Loops in the Disk Contract While Fixing the Basepoint

Proof. Let \(\gamma:[0,1]\to D^2\) be any loop based at \(1\). Define \(H:[0,1]\times[0,1]\to D^2\) by \(H(s,u)=(1-u)\gamma(s)+u\). The formula is continuous. For every \(s,u\), the triangle inequality gives \(|H(s,u)|\leq (1-u)|\gamma(s)|+u|1|\leq (1-u)+u=1\), so \(H(s,u)\in D^2\). Since \(\gamma(0)=\gamma(1)=1\), we have \(H(0,u)=H(1,u)=(1-u)1+u=1\) for all \(u\). Also \(H(s,0)=\gamma(s)\), and \(H(s,1)=1\). Thus \(H\) is a homotopy through based loops from \(\gamma\) to the constant loop. Every based loop has the same class as the constant loop, which proves the assertion about \(\pi_1(D^2,1)\). \(\square\)

This homotopy works because the disk is convex: each point on the straight segment from \(\gamma(s)\) to \(1\) remains in the disk. Fixing the basepoint is essential here. A contraction that moved the loop’s endpoint would not, by itself, establish that its class in the based fundamental group is the identity.

Induced Maps Respect Composition

Proof. For any based loop \(\alpha\) in \(X\), the definition gives \((g\circ f)_*([\alpha])=[(g\circ f)\circ\alpha]\). Associativity of composition gives \((g\circ f)\circ\alpha=g\circ(f\circ\alpha)\), and the definition of the induced maps gives \([g\circ(f\circ\alpha)]=g_*([f\circ\alpha])=g_*(f_*([\alpha]))\). These equalities hold for every class \([\alpha]\), proving the composition formula. For the identity map, \(\operatorname{id}_X\circ\alpha=\alpha\) for every loop, so \((\operatorname{id}_X)_*([\alpha])=[\alpha]\) for every class. The induced map is therefore the identity. \(\square\)

The induced map is well-defined on homotopy classes: if two based loops are homotopic, composing that homotopy with a continuous based map gives a homotopy of their image loops, with the basepoint fixed. This also explains why the calculation depends on the class of a loop and not on the particular loop chosen to represent it.

The No-Retraction Theorem

Proof. Suppose, for contradiction, that such a continuous map \(r\) exists. Let \(i:S^1\to D^2\) be the inclusion. The retraction condition says \(r\circ i=\operatorname{id}_{S^1}\), and \(r(1)=1\), so all three maps induce maps on fundamental groups with the chosen basepoints. By the composition proposition, \(r_*\circ i_*=(r\circ i)_*=(\operatorname{id}_{S^1})_*\), which is the identity map on \(\pi_1(S^1,1)\).

But \(\pi_1(D^2,1)\) has only its identity element, by the contraction proposition. Therefore \(i_*\) sends every element of \(\pi_1(S^1,1)\) to the identity of \(\pi_1(D^2,1)\). Applying \(r_*\) to that identity gives the identity of \(\pi_1(S^1,1)\), so \(r_*\circ i_*\) sends every class to the identity. This cannot be the identity map on \(\pi_1(S^1,1)\), because \([\alpha]\), represented by \(\alpha(s)=e^{2\pi i s}\), is nontrivial. We have a contradiction, so no such retraction exists. \(\square\)

What the Theorem Does—and Does Not—Rule Out

The theorem rules out a map that fixes every point of the boundary. It does not rule out every continuous map from the disk to the circle. For instance, the constant map \(F:D^2\to S^1\) given by \(F(z)=1\) is continuous and has values in \(S^1\). Its restriction to the boundary is constant, not the identity. At the boundary point \(-1\), \(F(-1)=1\ne-1\), so it is not a retraction.

Common pitfall. It is incorrect to argue that \(i_*\) is the identity because \(i\) leaves the points of \(S^1\) unchanged. The inclusion has codomain \(D^2\), and homotopies used to compare loops are allowed to move through that larger space. In fact, for the generator \([\alpha]\), the calculation is \(i_*([\alpha])=1_{\pi_1(D^2,1)}\), not \([\alpha]\): the loop contracts in the disk. What is the identity is the composite map \(r\circ i\), if a retraction exists. Keeping the domains and codomains visible prevents confusing these two claims.

A related tempting formula is \(z\mapsto z/|z|\), which sends every nonzero point of the disk to the circle and fixes each point on the boundary. It does not define a map on all of \(D^2\), because it is undefined at \(z=0\). Assigning a value at \(0\) does not repair continuity: along the positive real radius \(z=t\), \(t>0\), the formula equals \(1\), while along the negative real radius \(z=-t\), it equals \(-1\). These approach different values as \(t\) tends to \(0\). The no-retraction theorem shows that this failure is not merely a flaw in that particular formula: no other continuous formula can satisfy the boundary-fixing condition.

The argument is a useful model for an obstruction proof. A proposed map would force a relation between induced maps; the source and target fundamental groups make that relation impossible. Here, the inclusion destroys the circle’s nontrivial loop class when the loop is viewed in the disk, but a retraction would have to restore that class while composing with the inclusion. The mismatch is precisely what rules out the retraction.