What Must Be Checked in a Fundamental-Group Proof?
Many errors in fundamental-group arguments are not difficult calculations; they are failures to check whether the objects in a formula have the required endpoints or base points. A path concatenation is defined only when the endpoint of the first path is the starting point of the second. A loop homotopy used to identify elements of one fundamental group must keep the base point fixed. And a map induces a homomorphism between fundamental groups at specified points only if it sends the source base point to the target base point.
We make these checks explicit. Throughout, \(I=[0,1]\). A path in a topological space \(X\) is a continuous map \(\alpha:I\to X\); its starting point is \(\alpha(0)\), and its endpoint is \(\alpha(1)\). A loop based at \(x_0\in X\) is a path \(\alpha\) with \(\alpha(0)=\alpha(1)=x_0\). A path homotopy relative to endpoints from \(\alpha\) to \(\beta\) is a continuous map \(H:I\times I\to X\) such that \(H(0,t)=\alpha(t)\), \(H(1,t)=\beta(t)\), and \(H(s,0)=\alpha(0)=\beta(0)\), \(H(s,1)=\alpha(1)=\beta(1)\) for every \(s,t\in I\).
This convention fixes the order: \([\alpha][\beta]\) means travel along \(\alpha\), then along \(\beta\). The standard construction of the fundamental group shows that this product is well-defined on homotopy classes and gives a group; its associativity uses a reparameterization of paths. In the checks below, we will verify the endpoint and homotopy conditions relevant to each claim rather than treating the symbols as formal expressions.
Check Endpoints Before Concatenating
Worked Example: A Valid Concatenation, and Its Invalid Reversal
Let \(X=[0,2]\) with the subspace topology from \(\mathbb{R}\), and define \(\alpha(t)=t\) and \(\beta(t)=1+t\) for \(t\in I\). Their values lie in \(X\): \(0\le t\le1\) gives \(0\le\alpha(t)\le1\), and \(1\le1+t\le2\) gives \(1\le\beta(t)\le2\). The endpoints are \(\alpha(0)=0\), \(\alpha(1)=1\), \(\beta(0)=1\), and \(\beta(1)=2\). Thus \(\alpha(1)=\beta(0)\), so \(\alpha*\beta\) is defined.
Substituting into the concatenation formula gives $$ (\alpha*\beta)(t)= \begin{cases} 2t,&0\le t\le\frac12,\\ 1+(2t-1)=2t,&\frac12\le t\le1. \end{cases} $$ So \(\alpha*\beta\) is the path \(t\mapsto2t\) from \(0\) to \(2\). At the joining parameter \(t=\frac12\), both pieces equal \(1\), as required.
The reversed expression \(\beta*\alpha\) is not defined: \(\beta(1)=2\), but \(\alpha(0)=0\), so the endpoint of \(\beta\) does not match the starting point of \(\alpha\). The formula cannot be repaired by merely reversing the order of the symbols. Also note that \(\alpha,\beta\) are paths, not loops, so their classes do not give elements of a fundamental group at a common base point.
Check: The original choice \(X=I\), \(\alpha(t)=t\), \(\beta(t)=1+t\) would fail even earlier: \(\beta(1)=2\notin I\). It is not a path into \(I\), so its concatenation with \(\alpha\) cannot be a path in \(I\).
For loops based at the same point, both orders of concatenation are defined. That does not mean the two products are equal: it means only that each is a valid expression. A proof of equality must establish a homotopy or use a previously proved group relation. For general paths, even the existence of the reversed-order expression must first be checked.
Proof. The first formula for \(\alpha*\beta\) gives \(\alpha(1)\) at \(t=\frac12\), while the second gives \(\beta(0)\) there. If \(\alpha(1)\ne\beta(0)\), the two pieces do not agree at the joining parameter. Hence they do not define the stated concatenated path. \(\square\)
Fixed Base Points in Loop Homotopies
To represent equality in \(\pi_1(X,x_0)\), a homotopy must be a homotopy of loops based at \(x_0\) throughout. In formulas, in addition to \(H(0,t)=\alpha(t)\) and \(H(1,t)=\beta(t)\), one needs \(H(s,0)=H(s,1)=x_0\) for every \(s\in I\). It is not enough that the two endpoint loops happen to have the same base point.
Worked Example: A Homotopy with a Moving Base Point
Give \(S^1=\{z\in\mathbb{C}:|z|=1\}\) the subspace topology from \(\mathbb{C}\), and define \(H:I\times I\to S^1\) by \(H(s,t)=e^{2\pi i s}\). This is continuous and is constant as a function of \(t\) for each fixed \(s\). At \(s=0\), it is the constant loop at \(1\); at \(s=1\), it is also the constant loop at \(1\), since \(e^{2\pi i}=1\).
However, the intermediate loop at \(s=\frac12\) is constant at \(e^{\pi i}=-1\). In particular, \(H(\frac12,0)=H(\frac12,1)=-1\ne1\). This homotopy does not keep the base point \(1\) fixed. It therefore cannot, by itself, be cited as a based homotopy in \(S^1\) from the constant loop at \(1\) to itself. The example has equal endpoint loops, but that fact does not remove the requirement on the intermediate loops.
Check: For every \(s\), \(H(s,0)=H(s,1)=e^{2\pi i s}\), so each slice is indeed a loop. The failure is specifically that this common endpoint depends on \(s\), rather than remaining equal to the fixed base point \(1\).
Proof. Fix any \(s\in I\). The path \(t\mapsto H(s,t)\) starts at \(H(s,0)=x_0\) and ends at \(H(s,1)=x_0\), so it is a loop based at \(x_0\). The map \(H\) is continuous by assumption and its endpoint slices are \(H(0,\cdot)\) and \(H(1,\cdot)\). The stated endpoint condition is precisely what makes this a homotopy through loops based at the same point. \(\square\)
Induced Maps: The Base Point and the Product
A continuous map does not automatically give a map between fundamental groups at any two base points one chooses. To define the direct induced map, the target base point must be the image of the source base point.
Proof. First, if \(\alpha\) is a loop based at \(x_0\), then \(f\circ\alpha\) starts and ends at \(f(x_0)\), so it is a loop based at the specified target point. If \(\alpha\) and \(\beta\) are based-homotopic through \(H\), then \(f\circ H\) is continuous, has endpoint slices \(f\circ\alpha\) and \(f\circ\beta\), and satisfies \((f\circ H)(s,0)=(f\circ H)(s,1)=f(x_0)\). Thus it is a based homotopy, so the rule does not depend on the representative loop. Finally, whenever \(\alpha,\beta\) are loops based at \(x_0\), the two paths \(f\circ(\alpha*\beta)\) and \((f\circ\alpha)*(f\circ\beta)\) agree at every parameter: on \(0\le t\le\frac12\), both equal \(f(\alpha(2t))\); on \(\frac12\le t\le1\), both equal \(f(\beta(2t-1))\). Therefore $$ f_*([\alpha][\beta]) =[f\circ(\alpha*\beta)] =[(f\circ\alpha)*(f\circ\beta)] =f_*([\alpha])f_*([\beta]). $$ This is the homomorphism property. \(\square\)
Worked Example: The Base Point of a Map of the Circle
Give \(S^1\) the subspace topology from \(\mathbb{C}\), and let \(r:S^1\to S^1\) be \(r(z)=-z\). The map is continuous, and \(r(1)=-1\). Thus the direct induced map from loops based at \(1\) has target \(\pi_1(S^1,-1)\), not \(\pi_1(S^1,1)\).
For the constant loop \(c(t)=1\), the composite is \((r\circ c)(t)=-1\). It is a loop based at \(-1\), but it is not a loop based at \(1\). Consequently, writing \(r_*([c])\) as an element of \(\pi_1(S^1,1)\) by the direct formula would be a base-point error. A separate base-point-change construction can relate groups at different points, but it is not part of the direct definition of \(r_*\).
Check: \(c(0)=c(1)=1\), while \((r\circ c)(0)=(r\circ c)(1)=-1=r(1)\). The image has exactly the base point required for the target group \(\pi_1(S^1,-1)\).
Worked Example: A Square Map and Concatenation
Again use \(S^1\) with its subspace topology from \(\mathbb{C}\), and define \(f:S^1\to S^1\) by \(f(z)=z^2\). This is continuous and based at \(1\), since \(f(1)=1\). Consider the loop \(\sigma(t)=e^{2\pi i t}\), which starts and ends at \(1\). Direct substitution gives $$ (f\circ\sigma)(t)=(e^{2\pi i t})^2=e^{4\pi i t}. $$
On the other hand, for \(0\le t\le\frac12\), \((\sigma*\sigma)(t)=\sigma(2t)=e^{4\pi i t}\). For \(\frac12\le t\le1\), \((\sigma*\sigma)(t)=\sigma(2t-1)=e^{2\pi i(2t-1)}=e^{4\pi i t-2\pi i}=e^{4\pi i t}\), because \(e^{-2\pi i}=1\). Hence \(f\circ\sigma=\sigma*\sigma\) as paths, not merely as homotopy classes, and \(f_*([\sigma])=[\sigma][\sigma]\).
Check: At \(t=0\), both formulas give \(1\). At \(t=1\), both give \(e^{4\pi i}=1\). At the joining parameter \(t=\frac12\), the two concatenation pieces give \(\sigma(1)=1\) and \(\sigma(0)=1\), so the concatenation is continuous there.
Composition Order and a Common Mistake
Induced maps also respect composition, but their order follows the order of function composition. If \(f:X\to Y\) and \(g:Y\to Z\) are continuous, with base points \(x_0\), \(y_0=f(x_0)\), and \(z_0=g(y_0)\), then \((g\circ f)_*\) and \(g_*\circ f_*\) have the same domain and target.
Proof. Let \([\alpha]\in\pi_1(X,x_0)\). By the definition of each induced map, $$ (g\circ f)_*([\alpha])=[(g\circ f)\circ\alpha] =[g\circ(f\circ\alpha)] =g_*([f\circ\alpha]) =(g_*\circ f_*)([\alpha]). $$ The equality holds for every class in the domain, so the two maps are equal. The base points match because \(g(f(x_0))=z_0\). \(\square\)
Common pitfall. A tempting but incorrect calculation is \(f_*([\alpha][\beta])=[(f\circ\beta)*(f\circ\alpha)]\), reversing the concatenation order. The definition instead gives $$ f_*([\alpha][\beta]) =[f\circ(\alpha*\beta)] =[(f\circ\alpha)*(f\circ\beta)]. $$ The first segment is the image of \(\alpha\), because \(\alpha\) is traversed first. The reversal is not justified by applying \(f\), and may not even be defined if the endpoint of \(f\circ\beta\) does not match the starting point of \(f\circ\alpha\).
Before accepting a fundamental-group calculation, check the following items: the maps really take values in the stated spaces; each concatenation has matching endpoints; every loop in a claimed group is based at that group's specified point; any homotopy used to equate based loops fixes the base point throughout; and the product order in an induced-map calculation agrees with the chosen definition of concatenation. These checks are small, but each prevents a different invalid inference.
Check Your Understanding
For each item, write down the relevant endpoints or base points before deciding whether the claim is valid.
- If \(\alpha\) goes from \(a\) to \(b\) and \(\beta\) goes from \(b\) to \(c\), which of \(\alpha*\beta\) and \(\beta*\alpha\) is defined from the endpoint condition alone?
- What condition must a homotopy \(H:I\times I\to X\) satisfy to be a homotopy through loops based at \(x_0\)?
- If \(f:X\to Y\) is continuous, what must the target base point be for the direct induced map from \(\pi_1(X,x_0)\)?
- For \(f(z)=z^2\) on \(S^1\) and \(\sigma(t)=e^{2\pi i t}\), compute \(f\circ\sigma\) and compare it with \(\sigma*\sigma\) on each half of \(I\).
- For continuous based maps \(f:X\to Y\) and \(g:Y\to Z\), explain why the induced-map identity is \((g\circ f)_*=g_*\circ f_*\), not the reverse composition.