From a Product to Two Cones
A cone has one base and one point at its tip. A suspension has two tips: it starts with a product \(X\times I\), collapses the copy of \(X\) at one end to one point, and collapses the copy at the other end to a different point. The product between these ends remains, and it can be viewed as two cones meeting along their bases. We will define the quotient precisely and prove that this description gives a homeomorphism.
Let \((X,\mathcal{T})\) be any topological space, and let \(I=[0,1]\) have its standard subspace topology from \(\mathbb{R}\). Give \(X\times I\) the product topology. Recall that a quotient topology is defined using a surjection \(q:Z\to Y\): a subset \(V\subseteq Y\) is open exactly when \(q^{-1}(V)\) is open in \(Z\).
Thus every point strictly between the ends has its own class: if \(0<t<1\), then \(q_\Sigma(x,t)=q_\Sigma(y,s)\) holds exactly when \(x=y\) and \(t=s\). The classes \(S\) and \(N\) are distinct whenever \(X\ne\varnothing\). If \(X=\varnothing\), then \(X\times I=\varnothing\) and \(\Sigma X=\varnothing\); in particular, there are no endpoint classes in that case.
The Two-Cone Construction
We use the following cone convention. The cone \(CX\) is the quotient of \(X\times I\) that collapses \(X\times\{1\}\) to one point and makes no other identifications. Its base is the image of \(X\times\{0\}\). Take two copies \(C_-X\) and \(C_+X\) of this cone. In their disjoint union, identify the base point represented by \((x,0)\) in \(C_-X\) with the base point represented by \((x,0)\) in \(C_+X\), for each \(x\in X\), and make no other identifications. Give the resulting space \(R\) the quotient topology from \(C_-X\bigsqcup C_+X\). This is the gluing of two cones along their bases by the identity on \(X\).
The disjoint union has its disjoint-union topology: a subset is open precisely when its intersection with each summand is open there. Denote by \(j_-\) and \(j_+\) the maps from the two cone copies into \(R\). The cone parameter \(s=0\) marks a base, and \(s=1\) marks the tip. With this convention, the lower cone runs from the south tip to the common base, while the upper cone runs from the common base to the north tip.
Proof. We construct continuous maps in both directions and check that they are inverses. Define \(P:X\times I\to R\) by \(P(x,t)=j_-q_-(x,1-2t)\) when \(0\le t\le 1/2\), and \(P(x,t)=j_+q_+(x,2t-1)\) when \(1/2\le t\le1\), where \(q_-\) and \(q_+\) are the cone quotient maps. At \(t=1/2\), both formulas give the common base point indexed by \(x\), so \(P\) is well-defined. At \(t=0\), its value is the lower tip, independent of \(x\); at \(t=1\), its value is the upper tip, also independent of \(x\). Thus \(P\) is constant on every equivalence class defining \(\Sigma X\).
On each of the closed sets \(X\times[0,1/2]\) and \(X\times[1/2,1]\), the formula for \(P\) is continuous: it is a product map with a continuous affine change of interval coordinate, followed by a quotient map and an inclusion into \(R\). The two restrictions agree on the intersection \(X\times\{1/2\}\). The finite closed pasting lemma therefore makes \(P\) continuous. In this use of the lemma, for any closed set \(F\subseteq R\), the preimage \(P^{-1}(F)\) intersects each closed slab in a set closed in that slab. Each such intersection is closed in \(X\times I\), and their finite union is \(P^{-1}(F)\), hence is closed.
Since \(P\) is constant on the classes of \(q_\Sigma\), it induces a map \(F:\Sigma X\to R\) with \(Fq_\Sigma=P\). This induced map is continuous: for every open \(V\subseteq R\), \(q_\Sigma^{-1}(F^{-1}(V))=P^{-1}(V)\) is open, and the definition of the quotient topology then says \(F^{-1}(V)\) is open.
For the reverse direction, define maps from the cone copies by \(g_-(q_-(x,s))=q_\Sigma(x,(1-s)/2)\) and \(g_+(q_+(x,s))=q_\Sigma(x,(1+s)/2)\). The corresponding maps on \(X\times I\) are continuous. Each is constant on the cone equivalence classes: when \(s=1\), the first maps every \(x\) to \(S\), and the second maps every \(x\) to \(N\). The quotient topology of each cone therefore gives continuous maps \(g_-:C_-X\to\Sigma X\) and \(g_+:C_+X\to\Sigma X\). At \(s=0\), both send the base point indexed by \(x\) to \(q_\Sigma(x,1/2)\). They agree on the points identified in forming \(R\), so they induce a continuous map \(G:R\to\Sigma X\). Its continuity follows from the quotient topology on \(R\), by the same preimage test used for \(F\).
For \(0\le t\le1/2\), the lower-cone parameter used by \(P\) is \(s=1-2t\), and \(g_-\) sends this parameter back to \((1-s)/2=t\). For \(1/2\le t\le1\), the upper-cone parameter is \(s=2t-1\), and \(g_+\) sends it back to \((1+s)/2=t\). Hence \(GF\) fixes every class in \(\Sigma X\). Conversely, on a lower-cone point with parameter \(s\), the two maps return the lower-cone parameter \(1-2((1-s)/2)=s\); on an upper-cone point they return \(2((1+s)/2)-1=s\). Thus \(FG\) fixes every point of \(R\), including the common base. The maps are continuous inverses, so they are a homeomorphism. If \(X=\varnothing\), both spaces are empty and the unique maps between them are inverse homeomorphisms. \(\square\)
Worked Quotients
Worked Example: The Suspension of a Singleton
Let \(X=\{a\}\) with its unique topology \(\{\varnothing,\{a\}\}\). The product \(X\times I\) has one point \((a,t)\) for each \(t\in I\). Neither endpoint identification merges distinct points, because each endpoint fiber contains only \((a,0)\) or \((a,1)\). Therefore the quotient map \(q_\Sigma\) is a bijection from \(X\times I\) to \(\Sigma X\).
The map \(h:I\to\Sigma X\) defined by \(h(t)=q_\Sigma(a,t)\) is a continuous bijection. It is a homeomorphism because for every \(V\subseteq\Sigma X\), the quotient definition says \(V\) is open exactly when \(q_\Sigma^{-1}(V)\) is open; under the bijection \(t\mapsto(a,t)\), this is exactly the usual open-set test in \(I\). The two endpoint classes are \(h(0)\) and \(h(1)\), and they are distinct. This checks that the suspension keeps the two ends separate.
Worked Example: A Two-Point Discrete Space
Let \(X=\{a,b\}\) with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\). In the product \(X\times I\), the classes are exactly \([a,0]=\{(a,0),(b,0)\}\), \([a,1]=\{(a,1),(b,1)\}\), and the singleton classes \(\{(a,t)\}\), \(\{(b,t)\}\) for each \(t\) with \(0<t<1\). This list follows directly from the three clauses in the equivalence relation: only points at the same endpoint are merged, and all other classes are singletons.
Now examine the two cones. In each cone, the two base points indexed by \(a\) and \(b\) remain distinct, while all points at parameter \(s=1\) form that cone’s single tip. Gluing the cones identifies the lower base point indexed by \(a\) with the upper base point indexed by \(a\), and does the same for \(b\). It does not identify the two base points with each other. The lower and upper tips also remain distinct. The resulting class list matches the suspension list: two tips, two common-base points, and two points for each interior parameter on the corresponding cone side. The theorem’s parameter formulas send the common base to \(t=1/2\), the lower tip to \(t=0\), and the upper tip to \(t=1\), confirming the match.
Check: at \(t=1/2\), the classes \(\{(a,1/2)\}\) and \(\{(b,1/2)\}\) are still distinct, exactly as the two glued base points are distinct. The gluing identifies bases point by point, not by collapsing the whole base to one point.
Worked Example: An Indiscrete Two-Point Space
Let \(X=\{a,b\}\) have the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). The quotient classes as a set are the same types as in the preceding example: there is one class at \(t=0\), one at \(t=1\), and one class for each pair \((x,t)\) with \(0<t<1\). The topology, however, differs.
A basis for the product topology on \(X\times I\) consists of sets \(U\times V\), where \(U\) is open in \(X\) and \(V\) is open in \(I\). Since \(U\) must be either \(\varnothing\) or \(X\), every open subset of the product is \(X\times V\) for some open \(V\subseteq I\), including the empty set. Such a set is saturated for the suspension relation: if it contains one point of an endpoint fiber, it contains the whole fiber, and interior classes are singletons.
Consequently the open subsets of \(\Sigma X\) are exactly the sets \(q_\Sigma(X\times V)\) for open \(V\subseteq I\). Indeed, their preimages are \(X\times V\), which are open; conversely, the preimage of any quotient-open set has the form \(X\times V\), and the quotient-open set is the image of that preimage. The underlying set has two distinct interior points at each \(t\in(0,1)\), but the open-set test does not distinguish their coordinates. This example shows why specifying the topology on \(X\), not only listing the quotient classes, matters.
Common Pitfall: Identifying the Two Ends
A common error is to collapse \(X\times\{0\}\) and \(X\times\{1\}\) to the same point. That is a different equivalence relation from the one defining \(\Sigma X\). For \(X=\{a\}\), the correct endpoint classes are \(\{(a,0)\}\) and \(\{(a,1)\}\), which are distinct. The incorrect relation would make them equal. On the two-cone side, another error is to leave the bases unglued: that produces two separate cone pieces rather than the space \(R\). In the correct construction, the bases are identified point by point, while the two tips remain separate.
The construction is useful because it converts one quotient description into a concrete gluing description. The parameter formulas in the proof also keep the roles of the points precise: \(t=0\) is the lower tip, \(t=1/2\) is the common base, and \(t=1\) is the upper tip. Reversing the cone parameter on the lower half is essential; without it, that half would run from its base to its tip instead of from the south tip to the common base.
Check Your Understanding
Use the equivalence relation and quotient topology in each question.
- For nonempty \(X\), which points of \(X\times I\) represent the two suspension points? Why are those points distinct?
- With the cone convention in this tutorial, what cone parameter corresponds to suspension height \(t=1/4\) on the lower half?
- Describe the equivalence classes in the suspension of a three-point discrete space.
- Why does gluing two cone bases identify points indexed by the same \(x\), but not the base points indexed by different elements of \(X\)?
- For the indiscrete two-point space, what is the preimage in \(X\times I\) of an open set in the suspension?