From a Product to a Suspension
A suspension is formed from \(X\times I\) by collapsing all of \(X\times\{0\}\) to one point and all of \(X\times\{1\}\) to another. The interior levels \(X\times\{t\}\), for \(0<t<1\), are not collapsed in this construction. To define a map on the suspension, a map on the product must assign the same value to points that the quotient identifies. The quotient topology then gives an exact test for continuity.
Let \((X,\mathcal{T})\) be a topological space and let \(I=[0,1]\) have its usual subspace topology from \(\mathbb{R}\). Give \(X\times I\) the product topology. Define an equivalence relation on \(X\times I\) by identifying all points of \(X\times\{0\}\) with one another, identifying all points of \(X\times\{1\}\) with one another, and making no other identifications between distinct points.
If \(X=\varnothing\), then \(X\times I=\varnothing\) and \(\Sigma X=\varnothing\); there are no suspension points. The conditions below on endpoint slices are then vacuous. For nonempty \(X\), each endpoint slice is a single equivalence class, and the two classes are distinct.
The Continuity Criterion
Proof. First suppose \(g:\Sigma X\to Y\) is continuous. The quotient map \(q\) is continuous by the definition of the quotient topology: for every open \(W\subseteq\Sigma X\), \(q^{-1}(W)\) is open. A composition of continuous maps is continuous, so \(F=g\circ q\) is continuous. Also, if \(x,x'\in X\), then \(q(x,0)=q(x',0)\), and therefore \(F(x,0)=g(q(x,0))=g(q(x',0))=F(x',0)\). The same reasoning at \(t=1\) shows that \(F\) is constant on that endpoint slice.
Conversely, suppose \(F:X\times I\to Y\) is constant on each endpoint slice. If two points of \(X\times I\) have the same image under \(q\), then either they are the same point or they belong to one of the two endpoint slices. In either case they have the same \(F\)-value. Thus the rule \(g(q(x,t))=F(x,t)\) assigns exactly one value to each point of \(\Sigma X\), and defines a function \(g:\Sigma X\to Y\). It is unique because every point of \(\Sigma X\) is \(q(x,t)\) for some \((x,t)\). To check continuity, let \(V\subseteq Y\) be open. Then \(q^{-1}(g^{-1}(V))=F^{-1}(V)\), which is open because \(F\) is continuous. By the definition of the quotient topology, \(g^{-1}(V)\) is open in \(\Sigma X\). Hence \(g\) is continuous. This also proves the stated equivalence. \(\square\)
When \(X\ne\varnothing\), the endpoint condition can be written as the existence of two values \(y_0,y_1\in Y\) such that \(F(x,0)=y_0\) and \(F(x,1)=y_1\) for every \(x\in X\). There is no requirement that \(y_0=y_1\): the two endpoint slices are different equivalence classes. The continuity test has two separate parts: first check that \(F\) is constant on each class being collapsed, and then check that \(F\) is continuous on the product.
Worked Example: Two Different Curves with the Same End Values
Let \(X=\{a,b\}\) have the discrete topology, and let \(Y=I\) with its usual topology. Define \(F:X\times I\to I\) by \(F(a,t)=t\) and \(F(b,t)=t^2\). Show that \(F\) induces a continuous map \(g:\Sigma X\to I\), and compare the values on the two interior levels at \(t=\frac12\).
The endpoint values at \(t=0\) are \(F(a,0)=0\) and \(F(b,0)=0^2=0\), so \(F\) is constant on \(X\times\{0\}\). At \(t=1\), \(F(a,1)=1\) and \(F(b,1)=1^2=1\), so it is constant on \(X\times\{1\}\). For each fixed point of the discrete space \(X\), the function of \(t\) is either \(t\) or \(t^2\), both continuous on \(I\). Since the two slices \(\{a\}\times I\) and \(\{b\}\times I\) are open in the product and cover it, the preimage of any open subset of \(I\) is a union of open preimages in those slices. Thus \(F\) is continuous on \(X\times I\).
The theorem gives a unique continuous \(g:\Sigma X\to I\) with \(g\circ q=F\). At \(t=\frac12\), the two values are \(F(a,\frac12)=\frac12\) and \(F(b,\frac12)=(\frac12)^2=\frac14\). These points are not identified by \(q\), since their parameter is strictly between 0 and 1, and their different images under \(g\) are therefore permitted.
Check: at the bottom endpoint both values are \(0\), and at the top endpoint both are \(1\), exactly as required. The two interior values are \(\frac12\) and \(\frac14\), confirming that constancy is required on the endpoint slices, not on every level.
Worked Example: A Real-Valued Map on the Suspension of an Interval
Give \(X=[-1,1]\) the subspace topology from \(\mathbb{R}\), and define \(F:X\times I\to\mathbb{R}\) by \(F(x,t)=xt(1-t)\). Verify that it induces a continuous real-valued map on \(\Sigma X\), and compute some values.
The formula is a polynomial in the real coordinates \(x\) and \(t\), so its restriction to \(X\times I\) is continuous. For every \(x\in X\), \(F(x,0)=x\cdot0\cdot(1-0)=0\), and \(F(x,1)=x\cdot1\cdot(1-1)=0\). Both endpoint slices therefore have the same constant value, zero. The criterion gives a unique continuous map \(g:\Sigma X\to\mathbb{R}\) satisfying \(g(q(x,t))=xt(1-t)\).
At \((1,\frac14)\), the value is \(1\cdot\frac14\cdot\frac34=\frac{3}{16}\). At \((1,\frac34)\), it is \(1\cdot\frac34\cdot\frac14=\frac{3}{16}\), so these two values agree. At \((-1,\frac14)\), the value is \(-1\cdot\frac14\cdot\frac34=-\frac{3}{16}\); at \((-1,\frac34)\), it is \(-1\cdot\frac34\cdot\frac14=-\frac{3}{16}\). The negative-\(x\) values are the negatives of the corresponding positive-\(x\) values.
Check: substituting \(t=0\) gives \(g(q(x,0))=0\), and substituting \(t=1\) gives \(g(q(x,1))=0\) for every \(x\in[-1,1]\). Thus the formula assigns one value to each collapsed endpoint class, as required.
A Suspension with Two Meridians
Worked Example: A Map from the Suspension of Two Points to the Circle
Let \(X=\{+,-\}\) have the discrete topology, and let \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\) have the subspace topology from \(\mathbb{R}^2\). Define \(F:X\times I\to S^1\) by \(F(+,t)=(\cos(\pi t),\sin(\pi t))\) and \(F(-,t)=(\cos(\pi t),-\sin(\pi t))\). Verify that it induces a continuous map from \(\Sigma X\).
For either choice of sign, the sum of the squares of the coordinates is \(\cos^2(\pi t)+\sin^2(\pi t)=1\), so the values lie in \(S^1\). Each coordinate function is continuous in \(t\); since \(X\) is discrete, the two restrictions together give a continuous map on the product. At \(t=0\), \(\cos(0)=1\) and \(\sin(0)=0\), so \(F(+,0)=F(-,0)=(1,0)\). At \(t=1\), \(\cos(\pi)=-1\) and \(\sin(\pi)=0\), so \(F(+,1)=F(-,1)=(-1,0)\). Thus \(F\) is constant on each endpoint slice, and the theorem gives a unique continuous \(g:\Sigma X\to S^1\).
At \(t=\frac12\), \(\cos(\frac{\pi}{2})=0\) and \(\sin(\frac{\pi}{2})=1\). Therefore \(F(+,\frac12)=(0,1)\) and \(F(-,\frac12)=(0,-1)\). These are distinct points on the circle and are not required to agree: only points on the same collapsed endpoint slice must have equal values.
Check: both middle values satisfy \(0^2+1^2=1\), and the endpoint values \((1,0)\) and \((-1,0)\) each satisfy \(u^2+v^2=1\). The endpoints agree across each slice, while the two different endpoint values remain distinct.
What the Criterion Does—and Does Not—Require
A useful way to apply the theorem is to separate its logical requirements. Given \(F:X\times I\to Y\), first check the identifications: for all \(x,x'\in X\), verify \(F(x,0)=F(x',0)\), and separately verify \(F(x,1)=F(x',1)\). If either test fails, no function \(g\) can satisfy \(F=g\circ q\), regardless of whether \(F\) is continuous. If both tests pass, there is a unique induced function \(g\); its continuity is then equivalent to continuity of \(F\). In particular, the condition is not that \(F(x,t)\) be independent of \(x\) for interior values of \(t\).
Common pitfall. Consider the discrete space \(X=\{a,b\}\), the target \(\mathbb{R}\), and the continuous function \(F(a,t)=0\), \(F(b,t)=1\) for every \(t\in I\). Continuity holds because the function is constant on each of the two open slices \(\{a\}\times I\) and \(\{b\}\times I\). But \(q(a,0)=q(b,0)\), whereas \(F(a,0)=0\ne1=F(b,0)\). Thus there is no function \(g:\Sigma X\to\mathbb{R}\) with \(F=g\circ q\). The correction is to check constancy across each entire endpoint slice, not merely continuity along each parameter path.
The reverse mistake is to demand that the values at the two ends be equal. For any nonempty \(X\), the function \(F(x,t)=t\) is continuous and has the constant value \(0\) on \(X\times\{0\}\) and the constant value \(1\) on \(X\times\{1\}\). It induces a continuous map \(\Sigma X\to I\), even though \(0\ne1\). The two endpoint slices are collapsed separately, not together.
The criterion is useful whenever a map on a quotient is easier to describe before the identifications are made. One may define the formula on the product, check its ordinary continuity there, and verify that it respects exactly the equivalence classes used to form the suspension. The quotient topology then supplies continuity of the induced map without requiring a separate neighborhood calculation at the suspension points.
Check Your Understanding
For each question, distinguish the endpoint identifications from the continuity test on the product.
- Suppose \(F:X\times I\to Y\) is continuous and \(F(x,0)\) is independent of \(x\), but \(F(x,1)\) is not. Can \(F\) induce a function on \(\Sigma X\)? Explain.
- For nonempty \(X\), can an induced map have different values at the two suspension points? Give a continuous product map that demonstrates your answer.
- Let \(X=\{a,b\}\) be discrete and define \(F(a,t)=t\), \(F(b,t)=t^3\). Verify the endpoint conditions and find the two values at \(t=\frac12\).
- In the circle-valued example, why must the values at \(t=\frac12\) for \(+\) and \(-\) agree with one another neither by continuity nor by the quotient relation?
- State the quotient-topology step that proves the induced map \(g\) is continuous when \(F=g\circ q\) is continuous.