What It Means to Contract a Space
A contraction continuously moves every point of a space to one chosen point. The word “continuously” refers to the whole motion at once: the points of the space and the time parameter together form the domain of a single continuous map. This condition is stronger than asking that the map at each fixed time be continuous.
Throughout, \(I=[0,1]\) has its standard subspace topology from \(\mathbb{R}\), and \(X\times I\) has the product topology. For a point \(p\in X\), write \(c_p:X\to X\) for the constant map \(c_p(x)=p\). A homotopy from \(f:X\to Y\) to \(g:X\to Y\) is a continuous map \(H:X\times I\to Y\) satisfying \(H(x,0)=f(x)\) and \(H(x,1)=g(x)\) for every \(x\in X\).
For each \(t\in I\), the map \(C_t:X\to X\) defined by \(C_t(x)=C(x,t)\) is called the time-\(t\) map. A contraction starts with \(C_0=\operatorname{id}_X\) and ends with \(C_1=c_p\). The definition places no requirement that \(C_t(p)=p\) at intermediate times; it specifies the beginning and end of the motion, not what happens to \(p\) during the motion. If \(X\) is empty, it has no point \(p\) to contract to, so it is not contractible under this definition.
Contractibility and the Identity Map
Proof. Suppose first that \(X\) is contractible. By definition, there are \(p\in X\) and a continuous map \(C:X\times I\to X\) with \(C(x,0)=x\) and \(C(x,1)=p\) for all \(x\). Since \(C(x,0)=\operatorname{id}_X(x)\) and \(C(x,1)=c_p(x)\), this map is a homotopy from \(\operatorname{id}_X\) to \(c_p\).
Conversely, suppose \(p\in X\) and \(\operatorname{id}_X\simeq c_p\). By the definition of homotopy, there is a continuous \(H:X\times I\to X\) with \(H(x,0)=\operatorname{id}_X(x)=x\) and \(H(x,1)=c_p(x)=p\) for every \(x\). These are exactly the conditions for a contraction of \(X\) to \(p\). Thus \(X\) is contractible. \(\square\)
The theorem identifies two descriptions of the same structure: the contraction is a homotopy, and its endpoint maps are the identity and a constant map. The chosen point is part of the data of a contraction. Reversing the time parameter converts a homotopy from \(c_p\) to \(\operatorname{id}_X\) into one from \(\operatorname{id}_X\) to \(c_p\): if \(H\) has the former endpoints, then \((x,t)\mapsto H(x,1-t)\) has the latter.
One useful consequence is that a contractible nonempty space is path-connected. Here a path from \(x\) to \(p\) means a continuous map \(\gamma:I\to X\) with \(\gamma(0)=x\) and \(\gamma(1)=p\). If \(C\) is a contraction, define \(\gamma_x(t)=C(x,t)\). The map \(i_x:I\to X\times I\), \(i_x(t)=(x,t)\), is continuous by the definition of the product topology, so \(\gamma_x=C\circ i_x\) is continuous. Its endpoints are \(x\) and \(p\). In particular, for any \(x,y\in X\), the path from \(x\) to \(p\) followed by the reverse of the path from \(y\) to \(p\) joins \(x\) to \(y\). This is a necessary consequence of contractibility, not a sufficient test in general.
Checking Contractions in Examples
Worked Example: A Nonempty Indiscrete Space
Let \(X\) be any nonempty set with the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\), and choose \(p\in X\). Define \(C:X\times I\to X\) by \[ C(x,t)= \begin{cases} x,&t=0,\\ p,&0<t\leq 1. \end{cases} \] We check the endpoints first. At \(t=0\), \(C(x,0)=x\); at \(t=1\), \(C(x,1)=p\). Thus the required endpoint equations hold for every \(x\in X\).
To check continuity, the only open subsets of the codomain \(X\) are \(\varnothing\) and \(X\). Their preimages are respectively \(\varnothing\) and \(X\times I\), both open in the domain. Hence \(C\) is continuous, and \(X\) is contractible to \(p\). This example includes every nonempty indiscrete space, regardless of its cardinality. The topology matters: the formula may change abruptly at \(t=0\), but continuity into an indiscrete space still holds because there are only two open sets to test.
Worked Example: Contracting the Sierpinski Space
Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Define \(C:X\times I\to X\) by \(C(a,t)=a\) for every \(t\), \(C(b,0)=b\), and \(C(b,t)=a\) for \(0<t\leq1\). At time \(0\), \(C(a,0)=a\) and \(C(b,0)=b\), so \(C_0=\operatorname{id}_X\). At time \(1\), both points map to \(a\), so \(C_1=c_a\).
It remains to verify joint continuity. It suffices to check the preimages of the open sets \(\varnothing,\{a\},X\). The preimages of \(\varnothing\) and \(X\) are \(\varnothing\) and \(X\times I\), which are open. For the remaining open set, \[ C^{-1}(\{a\})=(\{a\}\times I)\cup(X\times(0,1]). \] The set \(\{a\}\) is open in \(X\), and \((0,1]\) is open in the subspace \(I\), since \((0,1]=I\cap(0,2)\). Each product in this union is open in the product topology, so the union is open. Therefore \(C\) is continuous and contracts \(X\) to \(a\).
Check: the preimage contains \((a,0)\), as required by \(C(a,0)=a\), but does not contain \((b,0)\), as required by \(C(b,0)=b\). For every \(t>0\), it contains both \((a,t)\) and \((b,t)\), as required by the formula. Thus the preimage calculation agrees with all the specified values of \(C\).
Worked Example: A Two-Point Discrete Space Is Not Contractible
Let \(X=\{0,1\}\) have the discrete topology \(\mathcal{T}=\mathcal{P}(X)\). Suppose a contraction \(C:X\times I\to X\) to \(0\) existed. Fix \(x=1\) and define \(\gamma:I\to X\) by \(\gamma(t)=C(1,t)\). The map \(i_1(t)=(1,t)\) is continuous, so \(\gamma=C\circ i_1\) is continuous. Its endpoints would be \(\gamma(0)=1\) and \(\gamma(1)=0\).
The inclusion \(j:X\to\mathbb{R}\), \(j(0)=0\) and \(j(1)=1\), is continuous: for every open \(U\subseteq\mathbb{R}\), the preimage \(j^{-1}(U)\) is a subset of the discrete space \(X\), hence open. Thus \(j\circ\gamma:I\to\mathbb{R}\) would be continuous with endpoint values \(1\) and \(0\). By the intermediate value theorem it would take the value \(1/2\), but \(j\circ\gamma\) takes values only in \(\{0,1\}\). This contradiction rules out a contraction to \(0\). A contraction to \(1\) would give the same contradiction by fixing \(x=0\). Therefore \(X\) is not contractible.
A Common Pitfall: Continuous Slices Are Not Enough
For a continuous homotopy \(H:X\times I\to Y\), every time map \(H_t(x)=H(x,t)\) is continuous: it is the composition of \(H\) with the continuous map \(x\mapsto(x,t)\). The converse does not hold. Continuity must be checked for the map on the product, not only for each fixed time.
Common pitfall. On the discrete space \(X=\{0,1\}\), define the proposed motion \[ H(x,t)= \begin{cases} x,&0\leq t<1/2,\\ 0,&1/2\leq t\leq1. \end{cases} \] Every time map \(H_t:X\to X\) is continuous because every map from a discrete space is continuous. The endpoint maps are also \(\operatorname{id}_X\) and \(c_0\). It is nevertheless wrong to call \(H\) a contraction: the open set \(\{0\}\subseteq X\) has preimage \[ H^{-1}(\{0\})=(\{0\}\times I)\cup(\{1\}\times[1/2,1]). \] This set is not open in \(X\times I\). In particular, every product neighborhood of \((1,1/2)\) contains some \((1,t)\) with \(t<1/2\), and that point is not in the preimage. Thus \(H\) is not jointly continuous. The path argument in the previous example confirms that no different formula can provide a contraction of this discrete space.
Contractibility is useful because it says that the identity map can be continuously deformed all the way to a map that forgets every input and returns one point. The criterion is exact: to prove contractibility, give the point and the jointly continuous homotopy; to disprove it, show that no such homotopy can exist. Endpoint equations alone, or continuity of the time maps alone, do not establish the required contraction.
Check Your Understanding
For each question, distinguish the endpoint conditions from the joint-continuity condition.
- State the equations a contraction to \(p\) must satisfy at \(t=0\) and \(t=1\).
- Explain why a homotopy from \(\operatorname{id}_X\) to \(c_p\) gives a contraction, and why a contraction gives such a homotopy.
- For the Sierpinski-space contraction, compute \(C^{-1}(\{a\})\) and explain why it is open in the product topology.
- Why does the intermediate value theorem rule out a path from \(1\) to \(0\) in the two-point discrete space?
- Give one reason that checking continuity of every time map separately is not enough to establish a homotopy.