Deformation Retraction of a Cylinder Onto Its Core

The Cylinder and Its Core

A cylinder has a natural circle running halfway between its two boundary circles. To deform the whole cylinder onto that core, keep each point’s circular coordinate unchanged and move its height linearly toward the midpoint. The formula is simple, but a deformation retraction requires more than arriving at the core: the homotopy must be continuous, and every point of the core must remain fixed at every time.

Let \(S^1=\{z\in\mathbb{C}:|z|=1\}\), with the subspace topology from \(\mathbb{C}\) with its standard topology, and let \(I=[0,1]\) with its standard topology. Give \(X=S^1\times I\) the product topology. Its core is the subspace \(A=S^1\times\{1/2\}\). A point of \(X\) has the form \((z,s)\), where \(z\) records its position around the circle and \(s\) its height.

The last condition is the defining extra requirement of a strong deformation retraction: the entire subspace \(A\), not merely its points at the start and finish, stays fixed throughout the deformation.

The Coordinate Formula and Its Verification

Define the retraction \(r:X\to A\) and the proposed homotopy \(H:X\times I\to X\) by

The second coordinate of \(H\) is a weighted average of the original height \(s\) and the core height \(1/2\). At \(t=0\) the height is \(s\); at \(t=1\) it is \(1/2\). To verify the formula rigorously, first check that it takes values in \(X\). For \(s,t\in I\), both \(1-t\) and \(t\) are nonnegative, so \((1-t)s+t/2\ge 0\). Also, since \(s\le 1\),

Thus the height remains in \(I\), while the first coordinate remains the same \(z\in S^1\).

Proof. We verify continuity and each required condition. The coordinate maps of \(r\) are the projection \((z,s)\mapsto z\) and the constant map \((z,s)\mapsto 1/2\), so both are continuous. A map into a product is continuous when each coordinate map is continuous: the product topology has a basis of products of open sets, and the preimage of such a product is the intersection of the preimages under the coordinate maps. Therefore \(r\) is continuous. Its values lie in \(A\), and for every \((z,1/2)\in A\), \(r(z,1/2)=(z,1/2)\), so it is a retraction.

For \(H\), the first coordinate is the projection \(((z,s),t)\mapsto z\), hence is continuous. The second coordinate \(((z,s),t)\mapsto (1-t)s+t/2\) is a polynomial in the real coordinates \(s,t\), hence is continuous. The product-coordinate criterion gives continuity into \(S^1\times I\); the range check above ensures that this is indeed the target. Substitution gives \(H((z,s),0)=(z,s)\) and \(H((z,s),1)=(z,1/2)=i(r(z,s))\). Finally, if \((z,s)\in A\), then \(s=1/2\), and for every \(t\in I\), \((1-t)s+t/2=(1-t)/2+t/2=1/2\). Hence \(H((z,1/2),t)=(z,1/2)\) throughout the homotopy. All the defining conditions hold. \(\square\)

Worked Examples Along the Homotopy

What the Retraction Gives

The retraction and inclusion are homotopy inverses. This conclusion follows directly from the verified maps, with the domains of the compositions made explicit.

Proof. The composition \(r\circ i:A\to A\) satisfies \((r\circ i)(a)=r(a)=a\) for every \(a\in A\), so \(r\circ i=\operatorname{id}_A\). The other composition has domain and codomain \(X\): \(i\circ r:X\to X\). The homotopy \(H\) has \(H(x,0)=x=\operatorname{id}_X(x)\) and \(H(x,1)=i(r(x))=(i\circ r)(x)\), so \(H\) is a homotopy from \(\operatorname{id}_X\) to \(i\circ r\). Thus the two maps satisfy the homotopy-inverse conditions with the identity maps on the correct spaces. \(\square\)

This does not say that the cylinder and its core are homeomorphic. The retraction \(r\) is not injective: for any \(z\in S^1\), the distinct points \((z,0)\) and \((z,1)\) both map to \((z,1/2)\). The conclusion is that the cylinder can be continuously deformed onto its core while keeping the core fixed, and hence the inclusion and retraction are homotopy inverses.

A Common Formula Error

Common pitfall. Replacing the correct height \((1-t)s+t/2\) by \(s+t/2\) does not give a homotopy in the cylinder. For example, starting at height \(s=1\) and taking \(t=1\) gives \(1+1/2=3/2\), which is greater than \(1\). The proposed map therefore leaves the allowed height interval. It also fails to end at the core height.

The corrected expression \((1-t)s+t/2\) decreases the weight on the starting height as time increases and assigns the remaining weight to \(1/2\). Its weights \(1-t\) and \(t\) are nonnegative and sum to \(1\), so the height stays between \(s\) and \(1/2\). At \(t=0\) it is \(s\), at \(t=1\) it is \(1/2\), and when \(s=1/2\) it remains \(1/2\) for every \(t\). Each of these checks is necessary: the endpoint conditions alone do not guarantee that a proposed homotopy stays in the space or fixes the target subspace throughout.