Why Fixing the Base Point Matters
A loop is not just a path that returns to where it started: in the fundamental group, its chosen starting point is part of the data. A homotopy that contracts the loop but moves that point may fail to be a homotopy of loops based at the original point. When a contraction keeps the base point fixed, however, it gives a direct formula for a based null-homotopy.
Throughout, \(I=[0,1]\) has its standard topology, and products such as \(X\times I\) have the product topology. Let \((X,\mathcal{T})\) be a topological space. A loop based at \(p\in X\) is a continuous map \(\gamma:I\to X\) with \(\gamma(0)=\gamma(1)=p\). The constant loop at \(p\) is the map \(c_p(s)=p\).
The variable \(s\) runs along each loop, and \(t\) moves through the homotopy. The conditions at \(s=0\) and \(s=1\) are therefore not optional: they are what make the homotopy based. A homotopy from \(\gamma\) to \(c_p\) that does not keep these endpoints at \(p\) is not, by itself, a based null-homotopy.
The Explicit Construction
Proof. Define \(G:I\times I\to X\times I\) by \(G(s,t)=(\gamma(s),t)\). This map is continuous: for each basic open set \(U\times V\) in the product topology, \(G^{-1}(U\times V)=\gamma^{-1}(U)\times V\), which is open in \(I\times I\). Thus \(H=C\circ G\) is continuous. At \(t=0\), \(H(s,0)=C(\gamma(s),0)=\gamma(s)\). At \(t=1\), \(H(s,1)=C(\gamma(s),1)=p\). At either endpoint of the loop, \(\gamma(0)=\gamma(1)=p\), so \(H(0,t)=C(p,t)=p=H(1,t)\). These are all the required conditions for a based null-homotopy. \(\square\)
The proof uses the contraction only through its defining conditions. The formula does not require a special shape for the loop, and the continuity of \(C\) is what ensures that the entire two-parameter family varies continuously.
Worked Example: A Loop in the Unit Interval
Give \(X=I=[0,1]\) its standard subspace topology from \(\mathbb{R}\), and choose base point \(p=0\). Define \(C(x,t)=(1-t)x\). This is continuous, \(C(x,0)=x\), \(C(x,1)=0\), and \(C(0,t)=0\), so it contracts \(X\) to \(0\) while fixing \(0\).
Consider the loop \(\gamma(s)=s(1-s)\). Its values lie in \(I\), since \(0\leq s(1-s)\leq 1/4\) for \(0\leq s\leq1\), and \(\gamma(0)=\gamma(1)=0\). The theorem gives
For instance, \(H(1/2,0)=(1)(1/2)(1/2)=1/4\), the maximum value of the original loop. At the same position halfway through the homotopy, \(H(1/2,1/2)=(1/2)(1/2)(1/2)=1/8\). At the final time, \(H(1/2,1)=(0)(1/2)(1/2)=0\).
Check: For every \(t\), \(H(0,t)=(1-t)0(1-0)=0\) and \(H(1,t)=(1-t)1(1-1)=0\). Also \(H(s,0)=s(1-s)\) and \(H(s,1)=0\). Thus the formula begins at \(\gamma\), ends at the constant loop, and fixes the base point throughout.
Checking the Endpoint Condition
Proof. Since \(\gamma(0)=\gamma(1)=p\), the formula gives \(H(0,t)=C(p,t)=H(1,t)\) for each \(t\). If \(C(p,t)=p\) for every \(t\), both endpoints stay at \(p\). Conversely, if the endpoint condition holds, then in particular \(H(0,t)=p\) for every \(t\). But \(H(0,t)=C(p,t)\), so \(C(p,t)=p\) for every \(t\). \(\square\)
This proposition concerns the specific homotopy obtained by applying the contraction point by point to the loop. It does not say that a contraction moving \(p\) can never be used to obtain a based null-homotopy by some other construction. It says that the direct formula has the required endpoints exactly when the contraction fixes \(p\).
Worked Example: A Contracting Homotopy That Moves the Base Point
Let \(X=D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq1\}\) with the subspace topology from the Euclidean plane. Set \(p=(1,0)\), and define \(C((x,y),t)=((1-t)x,(1-t)y)\). This is a continuous contraction of the disk to \((0,0)\): at \(t=0\) it is the identity, and at \(t=1\) every point maps to \((0,0)\). But it does not fix \(p\), since \(C(p,t)=(1-t,0)\).
Take the loop \(\gamma(s)=(\cos(2\pi s),\sin(2\pi s))\), based at \(p\). Applying the direct formula gives \(H(s,t)=((1-t)\cos(2\pi s),(1-t)\sin(2\pi s))\). At the two endpoints of the loop,
At \(t=1/2\), both endpoints are \((1/2,0)\), not \(p=(1,0)\). At \(t=1\), the whole loop has become the constant loop at \((0,0)\), also not at \(p\). Thus this formula contracts the loop but does not give a based null-homotopy at \(p\).
Check: The failure occurs already at the base point: \(C(p,1/2)=((1-1/2)1,(1-1/2)0)=(1/2,0)\ne(1,0)\). The proposition predicts exactly this endpoint failure. A correction must account for the path traced by the base point during the contraction; the unadjusted formula does not do so.
Two More Checks on the Construction
Worked Example: A Loop in a Disk with Fixed Base Point
Again give \(D^2\) the Euclidean subspace topology, but now choose \(p=(0,0)\) and use the contraction \(C((x,y),t)=((1-t)x,(1-t)y)\). This contraction fixes \(p\). Consider \(\gamma(s)=(\tfrac12\sin(2\pi s),0)\). Because \(|\tfrac12\sin(2\pi s)|\leq\tfrac12\), this point lies in \(D^2\) for every \(s\), and \(\gamma(0)=\gamma(1)=(0,0)\).
The based null-homotopy is
At \(s=1/4\) and \(t=1/2\), \(\sin(2\pi(1/4))=\sin(\pi/2)=1\), so \(H(1/4,1/2)=(1/4,0)\). At \(s=0\), \(H(0,t)=(\tfrac12(1-t)\sin(0),0)=(0,0)\); at \(s=1\), \(H(1,t)=(\tfrac12(1-t)\sin(2\pi),0)=(0,0)\).
Check: At \(t=0\) the formula is exactly \(\gamma\), and at \(t=1\) its first coordinate is \(\tfrac12(0)\sin(2\pi s)=0\) for every \(s\). So the loop contracts to the constant loop at the same base point.
Worked Example: A Two-Point Indiscrete Space
Let \(X=\{p,q\}\) have the indiscrete topology \(\{\varnothing,X\}\), and choose base point \(p\). Define \(C:X\times I\to X\) by \(C(x,0)=x\) and \(C(x,t)=p\) for \(t>0\). Every map into an indiscrete space is continuous, because the only open sets in the codomain have preimages \(\varnothing\) or the whole domain. Hence \(C\) is continuous. It satisfies \(C(x,0)=x\), \(C(x,1)=p\), and \(C(p,t)=p\) for all \(t\).
Define \(\gamma(1/2)=q\) and \(\gamma(s)=p\) for every \(s\ne1/2\). This is continuous into \(X\) for the same reason, and it is a loop based at \(p\). The constructed homotopy is \(H(s,t)=C(\gamma(s),t)\): at \(t=0\) it equals \(\gamma\), and for \(t>0\) it is constantly \(p\).
Check: At \(s=0\) and \(s=1\), \(\gamma(s)=p\), so \(H(0,t)=H(1,t)=C(p,t)=p\). Although the formula changes abruptly at \(t=0\) when viewed just as a function of point labels, it is continuous in the stated topology because every map into \(X\) is continuous. This example makes the role of the topology explicit.
What the Construction Does—and Does Not—Show
When a contraction fixes \(p\), each based loop is based null-homotopic by a single composition. By the standard null-homotopy criterion for the fundamental group, such a loop represents the identity class. The key point here is the construction of the homotopy with its endpoints checked, not merely the fact that the loop’s image can be contracted as a set.
When the contraction moves \(p\), its path \(t\mapsto C(p,t)\) is the path followed by the endpoints in the direct formula. If that path is not constantly \(p\), the intermediate paths need not be loops based at \(p\). Additional care is therefore required before drawing a conclusion about based homotopy or fundamental-group classes. In particular, do not infer a based null-homotopy merely from a contraction whose endpoint moves.
Common mistake. In the disk example, one might write \(H(s,t)=C(\gamma(s),t)\) and claim that the endpoints remain at \(p\) because \(\gamma(0)=\gamma(1)=p\). Substitution shows the error: \(H(0,t)=C(p,t)=(1-t,0)\), which is not \(p=(1,0)\) when \(t>0\). The correction is to check \(C(p,t)\) itself. For the direct construction to be based, the contraction must fix \(p\) throughout.
Check Your Understanding
For each question, write out the endpoint conditions before deciding whether a homotopy is based.
- State the three boundary conditions that a based null-homotopy of a loop at \(p\) must satisfy.
- For \(C(x,t)=(1-t)x\) on \(I\) and \(p=0\), verify directly that \(C\) fixes the base point.
- For a loop \(\gamma\) at \(p\), compute \(H(0,t)\) when \(H(s,t)=C(\gamma(s),t)\). What condition on \(C\) makes it equal to \(p\) for all \(t\)?
- In the disk example with \(p=(1,0)\), calculate both endpoints at \(t=1/4\). Are they the original base point?
- Why does a contraction that moves \(p\) fail to justify the direct based-null-homotopy construction, even if its final map is constant?