Why a Moving Base Point Needs Attention
A contraction shrinks an entire space to one point. Applying it to a loop gives a family of loops, but their base point may move during the contraction. A homotopy that moves the base point is not, by itself, a homotopy of loops based at the original point. We will correct for this movement by adding the path traced by the base point and then retracing that path in reverse.
We use the following conventions. A loop based at \(x_0\in X\) is a continuous map \(\gamma:I\to X\), where \(I=[0,1]\), with \(\gamma(0)=\gamma(1)=x_0\). A homotopy of loops based at \(x_0\) is a continuous map \(K:I\times I\to X\) such that \(K(s,0)\) and \(K(s,1)\) are the loops at the beginning and end, and \(K(0,t)=K(1,t)=x_0\) for every \(t\in I\). The fundamental group \(\pi_1(X,x_0)\) consists of based-homotopy classes of loops, with concatenation as its operation; its identity is the class of the constant loop at \(x_0\).
The topology on \(X\times I\) in this definition is the product topology. The contraction is an unbased homotopy from the identity map of \(X\) to the constant map with value \(p\). If we begin with a loop \(\gamma\) based at \(x_0\), the paths \(s\mapsto H(\gamma(s),t)\) begin and end at \(H(x_0,t)\), not necessarily at \(x_0\). The path \(t\mapsto H(x_0,t)\) records exactly how the base point moves.
The Fixed-Basepoint Correction
Proof. Fix \(x_0\in X\), and let \(\gamma:I\to X\) be any loop based at \(x_0\). Choose a contraction \(H\) to a point \(p\), and define the basepoint track \(\alpha:I\to X\) by \(\alpha(t)=H(x_0,t)\). It is continuous, with \(\alpha(0)=x_0\) and \(\alpha(1)=p\). Define \(K:I\times I\to X\) by the following three formulas:
On the first region, \(K\) travels along the basepoint track from \(x_0\) to \(\alpha(t)\). On the middle region, it follows the contracted loop \(s\mapsto H(\gamma(s),t)\). On the last region, it retraces the track from \(\alpha(t)\) to \(x_0\). At \(s=\frac13\), the first formula gives \(\alpha(t)\), and the second gives \(H(\gamma(0),t)=H(x_0,t)=\alpha(t)\). At \(s=\frac23\), the second gives \(H(\gamma(1),t)=\alpha(t)\), and the third also gives \(\alpha(t)\). Thus the formulas agree on the overlaps. Each formula is continuous on its closed region, so the pasting lemma gives continuity of \(K\) on \(I\times I\).
For all \(t\), \(K(0,t)=\alpha(0)=x_0\) and \(K(1,t)=\alpha(0)=x_0\). Hence \(K\) is a homotopy through loops based at \(x_0\). At \(t=0\), the first and last parts are constant at \(x_0\), while the middle part traverses \(\gamma\). Consequently \(K(s,0)=\gamma(\rho(s))\), where \(\rho(s)=0\) on \([0,\frac13]\), \(\rho(s)=3s-1\) on \([\frac13,\frac23]\), and \(\rho(s)=1\) on \([\frac23,1]\). This is \(\gamma\) with constant pauses. It is based-homotopic to \(\gamma\): the map \((s,u)\mapsto\gamma((1-u)\rho(s)+us)\) is continuous and keeps both endpoints at \(x_0\), since the interpolated parameter is \(0\) at \(s=0\) and \(1\) at \(s=1\).
At \(t=1\), the middle part is constantly \(p\), since \(H(x,1)=p\) for every \(x\). The first and last parts travel along \(\alpha\) and its reverse, with a pause at \(p\). To verify that this loop is based null-homotopic, let \(\phi:I\to I\) equal \(3s\) for \(0\leq s\leq\frac13\), \(1\) for \(\frac13\leq s\leq\frac23\), and \(3(1-s)\) for \(\frac23\leq s\leq1\). Then \(K(s,1)=\alpha(\phi(s))\). The loop \(\alpha(\phi(s))\) is based-homotopic to the usual traversal of \(\alpha\) followed by its reverse: interpolate linearly between \(\phi(s)\) and the parameter \(h(s)\), where \(h(s)=2s\) for \(s\leq\frac12\) and \(h(s)=2-2s\) for \(s\geq\frac12\), and compose with \(\alpha\). The usual traversal is null-homotopic by the explicit based homotopy \((s,u)\mapsto\alpha((1-u)h(s))\); at \(u=0\) it traverses \(\alpha\) and then reverses it, and at \(u=1\) it is constantly \(x_0\). Thus \(K(s,1)\) is based null-homotopic. Since \(K\) connects a loop based-homotopic to \(\gamma\) to a null-homotopic loop, \(\gamma\) represents the identity in \(\pi_1(X,x_0)\). The loop \(\gamma\) was arbitrary, so every element of \(\pi_1(X,x_0)\) is the identity. The choice of \(x_0\) was arbitrary as well. \(\square\)
The proof supplies the essential correction: at each time \(t\), travel from \(x_0\) to the moving base point, traverse the contracted loop there, and return to \(x_0\). The resulting family is genuinely based, even though the contraction itself is not.
Worked Applications
Worked Example: The Real Line with a Moving Base Point
Give \(\mathbb{R}\) its standard topology and use \(H(x,t)=(1-t)x\). This is continuous, \(H(x,0)=x\), and \(H(x,1)=0\), so it contracts \(\mathbb{R}\) to \(0\). Choose the base point \(x_0=2\). Its track is \(\alpha(t)=2(1-t)\), which starts at \(\alpha(0)=2\) and ends at \(\alpha(1)=0\).
If \(\gamma\) is any loop based at \(2\), the uncorrected family is \(F(s,t)=(1-t)\gamma(s)\). Its endpoints are \(F(0,t)=F(1,t)=2(1-t)\). For example, at \(t=\frac12\), both endpoints are \(1\), not \(2\). Thus \(F\) is not a homotopy through loops based at \(2\).
The correction travels from \(2\) to \(2(1-t)\), follows \(F\), and returns from \(2(1-t)\) to \(2\). At \(t=1\), the middle part is constantly \(0\), while the two outer parts travel from \(2\) to \(0\) and back to \(2\). The theorem’s explicit cancellation homotopy contracts this resulting loop while fixing \(2\). Therefore \(\pi_1(\mathbb{R},2)\) is trivial. The theorem gives the same conclusion at every other real base point.
Check: substituting \(t=0\) into \(H\) gives \(H(x,0)=x\); substituting \(t=1\) gives \(H(x,1)=0\). The track’s endpoints are \(2(1-0)=2\) and \(2(1-1)=0\), as required.
Worked Example: A Two-Point Indiscrete Space
Let \(X=\{a,b\}\) have the indiscrete topology \(\{\varnothing,X\}\). Define \(H:X\times I\to X\) by \(H(x,0)=x\) and \(H(x,t)=a\) for \(t>0\). This map is continuous because the only open subsets of its codomain are \(\varnothing\) and \(X\), whose preimages are respectively \(\varnothing\) and \(X\times I\), both open. Also \(H(x,0)=x\) and \(H(x,1)=a\), so \(H\) is a contraction.
Take \(x_0=b\). The basepoint track has \(\alpha(0)=b\) and \(\alpha(t)=a\) for \(t>0\). It is continuous: every map into an indiscrete space is continuous, since the preimage of each open set is either empty or the entire domain. In particular, this track need not resemble a path in a Hausdorff space, and it still provides the correction used in the theorem. Hence \(\pi_1(X,b)\) is trivial. Applying the theorem with \(x_0=a\) also gives that \(\pi_1(X,a)\) is trivial.
Check: the contraction endpoint is \(H(a,1)=a\) and \(H(b,1)=a\), so the time-one map is constant. At time zero the map is the identity on both elements: \(H(a,0)=a\) and \(H(b,0)=b\).
Worked Example: An L-Shaped Subspace of the Plane
Give \(X=([0,1]\times\{0\})\cup(\{0\}\times[0,1])\) the subspace topology inherited from \(\mathbb{R}^2\) with its Euclidean topology. Define \(H:X\times I\to X\) by \(H((x,y),t)=((1-t)x,(1-t)y)\). This formula is continuous as the restriction of a continuous map on \(\mathbb{R}^2\times I\). It takes values in \(X\): if \(y=0\), the second coordinate remains zero; if \(x=0\), the first coordinate remains zero.
At \(t=0\), \(H((x,y),0)=(x,y)\). At \(t=1\), \(H((x,y),1)=(0,0)\). Thus \(H\) contracts \(X\) to \((0,0)\). Choose \(x_0=(1,0)\). The basepoint track is \(\alpha(t)=(1-t,0)\), with \(\alpha(0)=(1,0)\) and \(\alpha(1)=(0,0)\). The contraction does move this base point, so one must use the track correction rather than claim that applying \(H\) directly to a loop is a based homotopy. The theorem gives \(\pi_1(X,(1,0))\) trivial, and it gives a trivial fundamental group for every other point of \(X\) as well.
Check: for \((x,0)\in[0,1]\times\{0\}\), the image is \(((1-t)x,0)\), which remains on the horizontal segment. For \((0,y)\in\{0\}\times[0,1]\), the image is \((0,(1-t)y)\), which remains on the vertical segment. The two descriptions agree at \((0,0)\).
Consequences and a Common Pitfall
A nonempty contractible space is also path-connected. Indeed, for any \(x,y\in X\), the paths \(t\mapsto H(x,t)\) and \(t\mapsto H(y,t)\) run from \(x\) and \(y\) to the same point \(p\). Following the first path and then the reverse of the second gives a path from \(x\) to \(y\). Together with the theorem, this means a contractible space is simply connected: it is path-connected and has trivial fundamental group at every base point.
Common pitfall. In the real-line example, it would be wrong to say that \(F(s,t)=(1-t)\gamma(s)\) is a based homotopy of a loop at \(2\). Its endpoints equal \(2(1-t)\), which differs from \(2\) whenever \(t>0\). This calculation verifies that the base point moves; it does not show that the loop fails to become trivial. The repair is to include the track from \(2\) to \(2(1-t)\) before the loop and its reverse after the loop. The corrected family has endpoints \(2\) at every time, and the proof shows its final loop is null-homotopic.
The conclusion concerns based homotopy classes, not just the visual idea that a loop can be shrunk somewhere in the space. The track correction is what converts a contraction of the space into a valid based argument, and it works for every choice of base point.
Check Your Understanding
Use the definition of contraction and keep track of the base point in each argument.
- For the contraction \(H(x,t)=(1-t)x\) of \(\mathbb{R}\), compute the track of the base point \(x_0=-3\) and state its endpoints.
- In the L-shaped subspace, verify directly that scaling by \(1-t\) keeps a point on either segment inside that same segment.
- Why does the family \(H(\gamma(s),t)\) not automatically give a based homotopy when the contraction moves \(x_0\)?
- In the theorem’s corrected homotopy, what are the values at \(s=0\) and \(s=1\) for an arbitrary \(t\)?
- Explain why contractibility implies path-connectedness, including how to form a path between two specified points.