Computing Suspensions of Familiar Spaces

From a Product to a Suspension

A suspension starts with a space stretched along an interval, then collapses each end of that product to a point. This construction turns the two-point space \(S^0\) into a circle, and it turns a circle into a two-sphere. The same coordinate pattern works in every dimension: the suspension of \(S^n\) is \(S^{n+1}\). The quotient is important here: without collapsing the ends, the product is a cylinder, not a suspension.

Let \(I=[0,1]\) have its usual subspace topology from \(\mathbb{R}\). Give \(X\times I\) the product topology, and give its quotient the quotient topology. In this tutorial, \(S^n=\{u\in\mathbb{R}^{n+1}:\|u\|=1\}\) has the subspace topology inherited from Euclidean space.

Definition: For a topological space \(X\), its suspension \(\Sigma X\) is the quotient of \(X\times I\) by the equivalence relation that identifies all points \((x,0)\) with one another and identifies all points \((x,1)\) with one another, with no other identifications. The quotient map is denoted \(q:X\times I\to\Sigma X\). If \(X\ne\varnothing\), the classes of \(X\times\{0\}\) and \(X\times\{1\}\) are two distinct points, called the suspension points. If \(X=\varnothing\), then \(\Sigma X=\varnothing\).

For nonempty \(X\), points with interval coordinate strictly between \(0\) and \(1\) retain their \(X\)-coordinate: \(q(x,t)=q(y,s)\), with \(0<t,s<1\), holds exactly when \(x=y\) and \(t=s\). Each end, however, has just one class. In particular, the two suspension points are distinct because the equivalence relation never identifies points from opposite ends.

The Zero-Sphere Suspends to a Circle

Worked Example: The Suspension of \(S^0\)

The zero-sphere is \(S^0=\{-1,+1\}\subset\mathbb{R}\), with the subspace topology. Consider the map from \(S^0\times I\) to the unit circle \(S^1\subset\mathbb{R}^2\) given by \(g(u,t)=(u\sin(\pi t),-\cos(\pi t))\). Its values lie on \(S^1\), since for each \(u\in\{-1,+1\}\),

(u\sin(\pi t))^2+(-\cos(\pi t))^2 =\sin^2(\pi t)+\cos^2(\pi t)=1.

At \(t=0\), both values of \(u\) map to \((0,-1)\); at \(t=1\), both map to \((0,1)\). Thus the map respects exactly the endpoint identifications and induces a map \(\bar g:\Sigma S^0\to S^1\). For \(0<t<1\), the second coordinate \(-\cos(\pi t)\) determines \(t\), because \(\cos(\pi t)\) is strictly decreasing on \([0,1]\), and the first coordinate then determines \(u\), since \(\sin(\pi t)>0\). Every point of \(S^1\) other than \((0,-1)\) and \((0,1)\) has exactly one such \(t\) and one \(u\): its second coordinate is strictly between \(-1\) and \(1\), and its first coordinate is nonzero. The two endpoints each have one preimage class. Hence \(\bar g\) is a bijection.

The two choices \(u=-1\) and \(u=+1\) trace the left and right semicircles, respectively; their endpoints become the same two suspension points. To check a specific interior value, take \(u=+1\) and \(t=1/2\). Then \(\sin(\pi/2)=1\) and \(\cos(\pi/2)=0\), so the image is \((1,0)\), which satisfies \(1^2+0^2=1\). Below, the compact-to-Hausdorff argument proves that the induced bijection is a homeomorphism. Therefore \(\Sigma S^0\cong S^1\).

The Coordinate Map in All Dimensions

The preceding example suggests the general formula. A point \(u\in S^n\) is sent along a meridian: its first \(n+1\) coordinates shrink by the factor \(\sin(\pi t)\), while a new last coordinate moves from \(-1\) to \(1\). At \(t=0\) and \(t=1\), the first coordinates vanish, so the value does not depend on \(u\).

Theorem: For every integer \(n\geq 0\), \(\Sigma S^n\cong S^{n+1}\).

Proof. Define \(g:S^n\times I\to S^{n+1}\subset\mathbb{R}^{n+2}\) by \(g(u,t)=(\sin(\pi t)u,-\cos(\pi t))\). This map is continuous because its coordinate functions are continuous. Since \(\|u\|=1\), its values have norm one:

\|g(u,t)\|^2 =\sin^2(\pi t)\|u\|^2+\cos^2(\pi t) =\sin^2(\pi t)+\cos^2(\pi t)=1.

For every \(u\in S^n\), \(g(u,0)=(0,\ldots,0,-1)\) and \(g(u,1)=(0,\ldots,0,1)\). Thus \(g\) is constant on each equivalence class defining \(\Sigma S^n\), so it induces a map \(\bar g:\Sigma S^n\to S^{n+1}\) with \(\bar g(q(u,t))=g(u,t)\). It is continuous: for an open set \(V\subseteq S^{n+1}\), \(q^{-1}(\bar g^{-1}(V))=g^{-1}(V)\) is open, and the definition of the quotient topology says a subset of \(\Sigma S^n\) is open exactly when its preimage under \(q\) is open.

We show that \(\bar g\) is bijective. The two points of \(S^{n+1}\) whose last coordinate is \(-1\) or \(1\) are the images of the two suspension points. Any other point can be written \((v,s)\), where \(v\in\mathbb{R}^{n+1}\), \(-1<s<1\), and \(\|v\|^2+s^2=1\). There is a unique \(t\in(0,1)\) with \(s=-\cos(\pi t)\), because \(t\mapsto-\cos(\pi t)\) is strictly increasing from \(-1\) to \(1\). Then \(\sin(\pi t)>0\), and the equation \(\|v\|^2+s^2=1\) gives \(\|v\|=\sin(\pi t)\). Consequently \(u=v/\sin(\pi t)\) lies in \(S^n\), and \(g(u,t)=(v,s)\). This proves surjectivity. The same equations recover \(t\) uniquely from \(s\) and \(u\) uniquely from \(v\), so every such point has exactly one preimage in \(S^n\times(0,1)\). The two poles each have exactly one preimage class in the quotient. Thus \(\bar g\) is injective as well.

It remains to verify that this continuous bijection is a homeomorphism; a continuous bijection alone does not suffice. The set \(S^n\times I\) is a closed and bounded subset of \(\mathbb{R}^{n+2}\), so it is compact by the Heine–Borel theorem. Since \(q\) is continuous and onto, \(\Sigma S^n\) is compact. Also \(S^{n+1}\) is Hausdorff, as a subspace of Euclidean space. We use the following standard fact, with its proof: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Indeed, a closed subset of a compact space is compact, and its continuous image is compact. A compact subset \(K\) of a Hausdorff space is closed: for each \(y\notin K\), separate \(y\) and each \(x\in K\) by disjoint open neighborhoods; compactness gives finitely many neighborhoods covering \(K\), and the intersection of the corresponding neighborhoods of \(y\) is an open neighborhood disjoint from \(K\). Therefore the continuous bijection sends closed sets to closed sets, making its inverse continuous. Applying this fact to \(\bar g\) proves that \(\bar g\) is a homeomorphism. \(\square\)

Reading the Formula in Familiar Coordinates

Worked Example: Suspending the Circle

Write a point of \(S^1\) as \(u=(\cos\theta,\sin\theta)\), where \(\theta\in\mathbb{R}\). The coordinate map for its suspension is \((\cos\theta,\sin\theta,t)\mapsto(\sin(\pi t)\cos\theta,\sin(\pi t)\sin\theta,-\cos(\pi t))\). Its image lies in \(S^2\), because

\sin^2(\pi t)\cos^2\theta+\sin^2(\pi t)\sin^2\theta+\cos^2(\pi t) =\sin^2(\pi t)+\cos^2(\pi t)=1.

For example, \(\theta=\pi/3\) and \(t=1/2\) give \(\cos(\pi/3)=1/2\), \(\sin(\pi/3)=\sqrt{3}/2\), \(\sin(\pi/2)=1\), and \(\cos(\pi/2)=0\). The image is therefore \((1/2,\sqrt{3}/2,0)\), whose squared norm is \(1/4+3/4+0=1\). At \(t=0\) every \(\theta\) gives the south pole \((0,0,-1)\), and at \(t=1\) every \(\theta\) gives the north pole \((0,0,1)\). The theorem proves that these coordinates give a homeomorphism \(\Sigma S^1\cong S^2\).

Worked Example: Suspending the Two-Sphere

Use spherical coordinates \(u=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)\) on \(S^2\), where \(0\leq\phi\leq\pi\) and \(\theta\in\mathbb{R}\). Its suspension maps to \(S^3\subset\mathbb{R}^4\) by \((u,t)\mapsto(\sin(\pi t)\sin\phi\cos\theta,\sin(\pi t)\sin\phi\sin\theta,\sin(\pi t)\cos\phi,-\cos(\pi t))\). The squared norm is

\begin{aligned} &\sin^2(\pi t)\sin^2\phi\cos^2\theta +\sin^2(\pi t)\sin^2\phi\sin^2\theta\\ &\qquad+\sin^2(\pi t)\cos^2\phi+\cos^2(\pi t)\\ &=\sin^2(\pi t)\bigl(\sin^2\phi+\cos^2\phi\bigr)+\cos^2(\pi t)\\ &=\sin^2(\pi t)+\cos^2(\pi t)=1. \end{aligned}

For a numerical check, set \(\phi=\pi/2\), \(\theta=0\), and \(t=1/2\). The four coordinates are \((1,0,0,0)\), and their squared norm is \(1+0+0+0=1\). For either endpoint \(t=0\) or \(t=1\), the first three coordinates vanish, independently of \(\phi\) and \(\theta\). The general theorem gives \(\Sigma S^2\cong S^3\).

What the Computations Do—and Do Not—Say

The formula shows how the dimension increases: the original sphere coordinates are scaled by \(\sin(\pi t)\), and the extra coordinate records the position along the suspension. At intermediate values \(0<t<1\), that scale is positive; at the two ends it is zero, which is exactly why the whole end of the product can be collapsed to one point.

Common pitfall. One might claim that \(\Sigma S^0\) consists of two disjoint copies of \(I\), one for each point of \(S^0\). That is the product \(S^0\times I\), before taking the quotient. In the suspension, the two points \((-1,0)\) and \((+1,0)\) are identified, and \((-1,1)\) and \((+1,1)\) are identified. The resulting two arcs share both endpoints, as the explicit map to \(S^1\) verifies. The opposite ends are not identified with each other: their images are \((0,-1)\) and \((0,1)\), which are distinct.

The empty-space case is also worth keeping separate. If \(X=\varnothing\), then \(X\times I=\varnothing\), and its quotient is empty; there are no suspension points. The theorem about spheres is unaffected because \(S^n\) is nonempty for every \(n\geq0\).

Check Your Understanding

Use the quotient definition and, where appropriate, the coordinate formula to answer these questions.

For \(X\ne\varnothing\), why are the two suspension points distinct, and why does that assertion not apply to \(X=\varnothing\)?
In the formula for \(\Sigma S^n\), compute the image of any \(u\in S^n\) at \(t=0\) and at \(t=1\).
For a point \((v,s)\in S^{n+1}\) with \(-1<s<1\), explain how to recover its unique preimage class.
Why does a continuous bijection \(\Sigma S^n\to S^{n+1}\) become a homeomorphism in the proof?
Write the coordinate formula for suspending \(S^1\), and verify that its values have norm one.