From Based Loops to a Group
A loop records a way of leaving a point and returning to it. In topology, two loops may be treated as equivalent when one can be continuously deformed into the other without moving the base point or changing the endpoints during the deformation. The fundamental group packages these equivalence classes and uses loop concatenation as its operation. Both the space in which loops travel and the base point at which they begin are part of the construction.
Let \((X,\mathcal{T})\) be a topological space, let \(x_0\in X\), and give \(I=[0,1]\) its standard topology. A loop based at \(x_0\) is a continuous map \(\alpha:I\to X\) such that \(\alpha(0)=\alpha(1)=x_0\). Continuity is understood with respect to the standard topology on \(I\) and the topology \(\mathcal{T}\) on \(X\).
Here \(s\) is the deformation parameter and \(t\) is the loop parameter. For each fixed \(s\), the map \(t\mapsto H(s,t)\) is a loop based at \(x_0\). The conditions at \(t=0\) and \(t=1\) are essential: this is a homotopy through based loops, not merely a deformation of the images of two curves. The earlier material on endpoint-preserving homotopies establishes that this relation is an equivalence relation, so its equivalence classes partition the set of based loops.
The constant loop at \(x_0\), written \(c_{x_0}\), is defined by \(c_{x_0}(t)=x_0\) for every \(t\in I\). It is continuous and has both endpoints at \(x_0\), so it is one of the loops being classified. Thus \(\pi_1(X,x_0)\) is nonempty whenever \(x_0\in X\). If \(X\) has no points, there is no choice of base point and this construction does not apply.
Concatenation and the Intended Operation
If \(\alpha\) and \(\beta\) are loops based at \(x_0\), their concatenation, written \(\alpha*\beta\), traverses \(\alpha\) first and then \(\beta\). The following parametrization assigns half of the interval to each loop:
At the joining parameter \(t=\frac12\), the first formula gives \(\alpha(1)=x_0\), and the second gives \(\beta(0)=x_0\). The two pieces therefore agree there. The concatenation is a continuous loop based at \(x_0\), as established by the standard concatenation result for paths. The corresponding proposed operation on classes is
This formula expresses the intended group operation: follow a representative of the first class and then a representative of the second. There is a logical point to keep in view. A formula on representatives defines an operation on equivalence classes only if changing either representative leaves the resulting class unchanged. That well-definedness, and the group axioms, require proofs. The well-definedness of concatenation, associativity, the constant-loop identity, and the inverse law are established in the standard construction of the fundamental group. At this stage, the formula specifies the operation whose justification those results supply; it is not a substitute for those verifications.
Worked Example: Concatenating a Loop on the Circle
Give \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) the subspace topology inherited from the standard topology on \(\mathbb{R}^2\), and take \(p=(1,0)\) as base point. Define \(\gamma(t)=(\cos(2\pi t),\sin(2\pi t))\). Since \(\gamma(0)=\gamma(1)=(1,0)=p\), this is a loop based at \(p\).
Its concatenation with itself is given by \((\gamma*\gamma)(t)=\gamma(2t)\) on \(0\le t\le\frac12\), and by \((\gamma*\gamma)(t)=\gamma(2t-1)\) on \(\frac12\le t\le1\). Substituting the formula for \(\gamma\), these pieces are \((\cos(4\pi t),\sin(4\pi t))\) and \((\cos(4\pi t-2\pi),\sin(4\pi t-2\pi))\). They agree at \(t=\frac12\), where both give \((1,0)\).
Check: At \(t=0\), the value is \((\cos 0,\sin 0)=(1,0)\). At \(t=\frac14\), the first piece gives \((\cos\pi,\sin\pi)=(-1,0)\). At \(t=\frac12\), both pieces give \((1,0)\). At \(t=\frac34\), the second piece gives \((\cos\pi,\sin\pi)=(-1,0)\), and at \(t=1\) it gives \((\cos 2\pi,\sin 2\pi)=(1,0)\). Each value lies on \(S^1\), and the start and end are \(p\). Thus the formula gives a loop at the stated base point. It illustrates concatenation of representatives; it does not by itself determine whether this loop is equivalent to any other particular loop.
What the Space and Base Point Specify
The notation \(\pi_1(X,x_0)\) records two choices. First, each loop must take all its values in \(X\), and each homotopy must also take all its values in \(X\). Second, the endpoints of every loop and every intermediate loop in the homotopy must be the selected point \(x_0\). Consequently, changing either the space or the base point changes the defining collection of maps and the equivalence relation used to form classes.
This dependence should not be confused with a claim that every change produces a different group up to isomorphism. The notation records the input data; comparing the groups for different inputs requires additional arguments. In particular, a path between two base points does not, by itself, identify their fundamental groups. The construction of such comparisons is a separate result.
Worked Example: The Same Formula in Two Spaces
Let \(S^1\) have the subspace topology from \(\mathbb{R}^2\), and let \(\mathbb{R}^2\) have its standard topology. The loop \(\gamma(t)=(\cos(2\pi t),\sin(2\pi t))\) based at \(p=(1,0)\) is a map into \(S^1\). The same formula, with its codomain regarded as \(\mathbb{R}^2\), is also a based loop there.
In \(\mathbb{R}^2\), define \(H(s,t)=(1-s)\gamma(t)+sp\). This is continuous because its coordinate functions are sums and products of continuous real-valued functions. At \(s=0\), it is \(\gamma\); at \(s=1\), it is the constant loop at \(p\). For either endpoint \(t=0\) or \(t=1\), \(\gamma(t)=p\), so \(H(s,t)=(1-s)p+sp=p\) for every \(s\). Hence it is a based homotopy in \(\mathbb{R}^2\).
The same formula does not generally define a homotopy in \(S^1\). For instance, at \(s=\frac12\) and \(t=\frac14\), we have \(\gamma(\frac14)=(0,1)\), so \(H(\frac12,\frac14)=(\frac12,\frac12)\). Its squared distance from the origin is \((\frac12)^2+(\frac12)^2=\frac12\), not \(1\), so this value is not in \(S^1\). This only shows that this particular homotopy leaves \(S^1\); it does not prove that no other based homotopy in \(S^1\) exists. The example makes precise why homotopies must be required to remain in the specified space.
Worked Example: Two Base Points on the Circle
Give \(S^1\) its subspace topology, and choose \(p=(1,0)\) and \(q=(0,1)\). The loop \(\gamma_p(t)=(\cos(2\pi t),\sin(2\pi t))\) starts and ends at \(p\). The formula \(\gamma_q(t)=(\cos(\frac{\pi}{2}+2\pi t),\sin(\frac{\pi}{2}+2\pi t))\) starts and ends at \(q\), since its values at \(t=0\) and \(t=1\) are both \((0,1)\).
Thus \([\gamma_p]\) is a class in \(\pi_1(S^1,p)\), while \([\gamma_q]\) is a class in \(\pi_1(S^1,q)\). These symbols refer to classes formed using different endpoint conditions. A homotopy required to fix \(p\) at both ends cannot simultaneously satisfy the requirement to fix \(q\) at both ends when \(p\ne q\). This example identifies the role of the base point in the definition; it makes no claim that the two resulting groups are or are not isomorphic.
A Useful Result for Convex Subspaces
A simple setting gives a complete comparison of based-loop classes without needing any group calculations. A subset \(C\subseteq\mathbb{R}^n\) is convex if \((1-s)x+sy\in C\) for all \(x,y\in C\) and \(s\in I\). Give \(C\) the subspace topology from the standard topology on \(\mathbb{R}^n\).
Proof. Let \(\alpha,\beta:I\to C\) be based loops at \(x_0\). Define \(H:I\times I\to\mathbb{R}^n\) by \(H(s,t)=(1-s)\alpha(t)+s\beta(t)\). For each \((s,t)\), both \(\alpha(t)\) and \(\beta(t)\) belong to \(C\), and convexity implies \(H(s,t)\in C\). The coordinate functions of \(H\) are continuous, so \(H\), regarded as a map into the subspace \(C\), is continuous. For every \(t\), \(H(0,t)=\alpha(t)\) and \(H(1,t)=\beta(t)\). Since \(\alpha(0)=\alpha(1)=\beta(0)=\beta(1)=x_0\), for every \(s\) we have \(H(s,0)=(1-s)x_0+sx_0=x_0\) and \(H(s,1)=(1-s)x_0+sx_0=x_0\). Thus \(H\) is a based homotopy from \(\alpha\) to \(\beta\). Every pair of based loops represents the same equivalence class. There is at least one such loop, the constant loop, so the set of classes has exactly one element. \(\square\)
This theorem concerns the set of based-homotopy classes. It does not replace the separate verification that concatenation gives a well-defined group operation. It does show concretely that the chosen space controls which deformations are available: convexity supplies a straight-line homotopy staying inside the space.
Common Pitfall: Forgetting the Fixed Endpoints
A common error is to use a homotopy that moves the base point and then call it a based homotopy. For example, in \(\mathbb{R}\) with its standard topology, let \(c_0(t)=0\) and \(c_1(t)=1\). These are loops based at different points, so they are not both elements of the same based-loop set. The formula \(H(s,t)=s\) satisfies \(H(0,t)=c_0(t)\) and \(H(1,t)=c_1(t)\), but \(H(s,0)=s\), which is not fixed at \(0\) as \(s\) varies. It also has \(H(s,1)=s\), not a fixed endpoint. Therefore it is an ordinary homotopy between the two maps, not a based homotopy between loops at one base point.
The correction is to check all four boundary conditions in the definition: the two ends of the deformation must be the prescribed loops, and the values at the two loop endpoints must remain \(x_0\) throughout the deformation. When using \([\alpha]\cdot[\beta]=[\alpha*\beta]\), also remember that the representative-level formula alone does not establish independence from the representatives. That check is what allows the proposed operation to be an operation on classes rather than merely on loops.
Check Your Understanding
Use the definitions to answer the following questions.
- State the endpoint conditions that a homotopy must satisfy to be a based homotopy between loops at \(x_0\).
- For loops \(\alpha,\beta\) based at \(x_0\), which loop is traversed first in \(\alpha*\beta\)?
- Why does the formula \([\alpha]\cdot[\beta]=[\alpha*\beta]\) require a well-definedness proof?
- In the convex-subspace theorem, explain why the straight-line homotopy remains in \(C\) and keeps both loop endpoints fixed.
- Why does a based homotopy in \(\mathbb{R}^2\) not automatically count as a homotopy in \(S^1\)?