Radial Deformation Retraction of Punctured Euclidean Space

Moving Along Rays Without Reaching the Origin

Every nonzero vector has a direction and a positive length. To move a point of punctured Euclidean space onto the unit sphere, we can keep its direction and change its length continuously until that length is one. The important verification is that every intermediate point remains nonzero. We will make this radial motion into an explicit strong deformation retraction.

Fix an integer \(n\geq 1\). Give \(\mathbb{R}^n\) its standard Euclidean topology and norm, and give all subsets the subspace topology. Write \(X=\mathbb{R}^n\setminus\{0\}\) and \(S^{n-1}=\{x\in\mathbb{R}^n:\|x\|=1\}\subseteq X\). The parameter interval \(I=[0,1]\) has its standard topology, and \(X\times I\) has the product topology.

Here the candidate retraction sends each vector to the unit vector in its direction. To connect the identity map to this retraction, interpolate between the original vector and its normalized vector:

The formula is defined for every \(x\in X\), because \(\|x\|>0\). At the start, \(H(x,0)=x\); at the end, \(H(x,1)=x/\|x\|\). These endpoint computations alone do not prove that \(H\) works: we must also show that intermediate values never equal the removed origin, that the map is continuous, and that points on the sphere remain fixed.

The Radial Coordinate Calculation

Proof. Put \(r=\|x\|\), so \(r>0\), and put \(\lambda=(1-t)+t/r\). Since \(0\leq t\leq 1\), both \(1-t\) and \(t/r\) are nonnegative, and they cannot both be zero: \(1-t=0\) implies \(t=1\), in which case \(t/r=1/r>0\). Thus \(\lambda>0\). Since \(H(x,t)=\lambda x\), homogeneity of the Euclidean norm gives \(\|H(x,t)\|=\lambda\|x\|=((1-t)+t/r)r=(1-t)r+t\). This quantity is positive, so division by \(\|H(x,t)\|\) is valid. Therefore \(H(x,t)/\|H(x,t)\|=\lambda x/(\lambda r)=x/r\). The length changes, but the direction does not. \(\square\)

The length formula also directly rules out the main possible obstruction: \((1-t)\|x\|+t>0\), since it is a sum of nonnegative terms and cannot have both terms zero. Thus the path \(t\mapsto H(x,t)\) stays in \(X\), even if the starting length is less than one. The formula describes a linear change in radius from \(\|x\|\) to \(1\), not a linear change in the vector toward the origin.

Proof of the Strong Deformation Retraction

Proof. First, the formula for \(H\) is defined on all of \(X\times I\), since \(\|x\|>0\). The radial coordinate calculation shows that \(\|H(x,t)\|=(1-t)\|x\|+t>0\), so every value lies in \(X\). To check continuity, the norm map \(x\mapsto\|x\|\) is continuous, and \(u\mapsto1/u\) is continuous for \(u>0\). Consequently \(x\mapsto x/\|x\|\) is continuous on \(X\). The operations of scalar multiplication and addition in \(\mathbb{R}^n\) are continuous, so the displayed formula defines a continuous map from \(X\times I\) into \(\mathbb{R}^n\). Because its image lies in \(X\), it is also continuous as a map into the subspace \(X\).

At \(t=0\), the formula gives \(H(x,0)=x\); at \(t=1\), it gives \(H(x,1)=x/\|x\|=r(x)\), and \(\|r(x)\|=1\), so \(r(x)\in S^{n-1}\). For any \(a\in S^{n-1}\), \(\|a\|=1\), and hence \(H(a,t)=(1-t)a+ta=a\) for every \(t\in I\). In particular \(r(a)=a\), so \(r\) is a retraction, and the homotopy fixes the entire sphere pointwise. These are precisely the defining conditions for a strong deformation retraction. \(\square\)

The pointwise fixation condition matters: it is stronger than merely having the endpoint map land in the sphere. It says that the motion does not move any point that is already on the target. The formula achieves this because normalizing a unit vector does not change it.

Computing the Motion

The Homotopy-Type Consequence

A strong deformation retraction gives a homotopy equivalence. To state the consequence precisely, let \(i:S^{n-1}\to X\) be the inclusion \(i(a)=a\). Two maps are homotopic if a continuous map on the product with \(I\) joins them. A homotopy equivalence is a map with a map in the reverse direction such that both composites are homotopic to the corresponding identity maps.

Proof. The radial map \(r:X\to S^{n-1}\) and inclusion \(i:S^{n-1}\to X\) satisfy \(r\circ i=\operatorname{id}_{S^{n-1}}\), because \(r(a)=a\) for every \(a\in S^{n-1}\). The radial strong deformation retraction gives a homotopy from \(\operatorname{id}_X\) to \(i\circ r\): its initial map is \(H(x,0)=x\), and its final map is \(H(x,1)=i(r(x))\). Thus \(i\) and \(r\) are homotopy inverses, which proves the asserted homotopy equivalence. \(\square\)

This conclusion replaces the punctured space, which has points at every positive distance from the origin, by the sphere as a simpler representative of its homotopy type. The explicit construction also records exactly how the replacement happens: each point moves along its own ray, its radius changes linearly, and its direction stays constant.

A Common Mistake: Contracting Toward the Missing Point

A tempting but invalid construction is to use \(G(x,t)=(1-t)x\), because it moves points toward the origin. It does not define a homotopy in \(X\): for example, if \(x=(1,0)\in\mathbb{R}^2\setminus\{0\}\), then \(G((1,0),1)=(0,0)\), which is not in \(X\). So this formula leaves the required space at its endpoint and does not retract onto the unit circle.

The radial formula makes a different choice. For the same starting point, \(H((1,0),t)=((1-t)+t)(1,0)=(1,0)\), so it remains on the unit circle. For a point away from the circle, the scalar \((1-t)+t/\|x\|\) is always positive, as proved above. The correction is not merely to check the endpoints: every intermediate value must stay in the space on which the homotopy is defined.