Why Base Points Need Attention
A homotopy equivalence preserves the fundamental group, but the statement needs care: the maps in a homotopy inverse pair need not send the chosen base points to one another. The homotopies between their composites and the identity maps can also move base points. The resulting paths are not optional bookkeeping; they specify the base-point changes needed to form the inverse homomorphism.
For a space \(X\) and point \(x_0\in X\), \(\pi_1(X,x_0)\) is the group of homotopy classes of loops based at \(x_0\), with homotopies fixing the base point. We write paths in the order they are traversed: \(\gamma*\delta\) means first \(\gamma\), then \(\delta\), and \(\bar\gamma\) means the reverse path. A continuous map \(f:X\to Y\) with \(f(x_0)=y_0\) induces \(f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)\), given by \(f_*[\gamma]=[f\circ\gamma]\).
We will use two facts about base-point paths. If \(\lambda\) is a path from \(p\) to \(q\) in \(X\), define the base-point change isomorphism \(\lambda_\#:\pi_1(X,p)\to\pi_1(X,q)\) by \(\lambda_\#[\gamma]=[\bar\lambda*\gamma*\lambda]\). Also, if \(f:X\to Y\), then changing base point and applying \(f\) commute: \(f_*^q\circ\lambda_\#=(f\circ\lambda)_\#\circ f_*^p\). Here \(f_*^p\) and \(f_*^q\) use the base points \(p\) and \(q\), respectively. This identity follows by applying \(f\) to the three paths in \(\bar\lambda*\gamma*\lambda\). Base-point change is an isomorphism, with inverse \((\bar\lambda)_\#\).
The Track of a Homotopy
Proof. Let \(\gamma\) be a loop based at \(x_0\). The map \((s,t)\mapsto H(\gamma(s),t)\) gives a square: its bottom edge is \(F_0\circ\gamma\), its top edge is \(F_1\circ\gamma\), and its two side edges are both \(\lambda\), since \(\gamma(0)=\gamma(1)=x_0\). Traversing the bottom edge and then the right side is homotopic, with endpoints fixed, to traversing the left side and then the top edge. Thus \((F_0\circ\gamma)*\lambda\simeq\lambda*(F_1\circ\gamma)\). Prepending \(\bar\lambda\) gives \(\bar\lambda*(F_0\circ\gamma)*\lambda\simeq F_1\circ\gamma\), which is exactly \(\lambda_\#((F_0)_*[\gamma])=(F_1)_*[\gamma]\). This holds for every loop class. \(\square\)
The formula says that homotopic maps induce the same homomorphism only after the base point in the target has been matched using the homotopy’s track. If the track is a loop, the resulting base-point change acts by conjugation; it need not be the identity in an arbitrary fundamental group.
Homotopy Equivalences Give Isomorphisms
Proof. Choose homotopies \(H:g\circ f\simeq\operatorname{id}_X\) and \(K:f\circ g\simeq\operatorname{id}_Y\). Their tracks at the chosen points are \(\alpha(t)=H(x_0,t)\), from \(g(y_0)\) to \(x_0\), and \(\beta(t)=K(y_0,t)\), from \(f(g(y_0))\) to \(y_0\). The induced maps have different base points, so define \(\Phi=\alpha_\#\circ g_*:\pi_1(Y,y_0)\to\pi_1(X,x_0)\). By the identity and composition laws for induced maps, \((g\circ f)_*=g_*\circ f_*\), with the intermediate base points \(y_0\) and \(g(y_0)\). The homotopy track formula for \(H\) therefore gives \(\Phi\circ f_*=\alpha_\#\circ(g\circ f)_*=\operatorname{id}_{\pi_1(X,x_0)}\).
To check the other composition, first use the base-point naturality identity with \(f\) and \(\alpha\). It gives \[ f_*\circ\Phi =f_*^{x_0}\circ\alpha_\#\circ g_*^{y_0} =(f\circ\alpha)_\#\circ(f_*^{g(y_0)}\circ g_*^{y_0}) =(f\circ\alpha)_\#\circ(f\circ g)_*. \] The path \(f\circ\alpha\) and the path \(\beta\) have the same endpoints, both running from \(f(g(y_0))\) to \(y_0\). The homotopy track formula for \(K\) says \(\beta_\#\circ(f\circ g)_*=\operatorname{id}\), so \((f\circ g)_*=\beta_\#^{-1}\). Hence \(f_*\circ\Phi=(f\circ\alpha)_\#\circ\beta_\#^{-1}\), an isomorphism because both factors are base-point change isomorphisms. Set \(A=f_*\circ\Phi\). Since \(\Phi\circ f_*=\operatorname{id}\), we have \(A\circ A=f_*\circ(\Phi\circ f_*)\circ\Phi=A\). But \(A\) is invertible, so composing this equality with \(A^{-1}\) gives \(A=\operatorname{id}\). Thus \(f_*\circ\Phi=\operatorname{id}\) as well, and \(\Phi\) is the inverse of \(f_*\). \(\square\)
The two homotopy tracks need not be the same path. The proof does not assume that the two composites of the proposed inverse maps are immediately identity homomorphisms: the second composite is first identified as an isomorphism, and the left-inverse identity then forces it to be the identity too.
Worked Applications
Worked Example: Retracting a Cylinder onto Its Circle
Let \(X=S^1\times I\) and \(Y=S^1\), both with their usual topologies as subspaces of Euclidean spaces. Define \(f:X\to Y\) by \(f(z,t)=z\), and \(g:Y\to X\) by \(g(z)=(z,0)\). Take \(x_0=(1,0)\), so \(y_0=f(x_0)=1\). Then \(g(f(z,t))=(z,0)\), and \(f(g(z))=z\).
A homotopy from \(g\circ f\) to \(\operatorname{id}_X\) is \(H((z,t),s)=(z,st)\): at \(s=0\) it is \((z,0)\), and at \(s=1\) it is \((z,t)\). Its track at \(x_0\) is constant, since \(H((1,0),s)=(1,0)\). The other composite \(f\circ g\) is already the identity, with constant homotopy and constant track at \(1\). Thus the inverse on fundamental groups is simply \(g_*\), with no nontrivial base-point adjustment. Indeed \(f_*\circ g_*=(f\circ g)_*=\operatorname{id}\) and \(g_*\circ f_*=(g\circ f)_*=\operatorname{id}\). The theorem gives \(\pi_1(S^1\times I,(1,0))\cong\pi_1(S^1,1)\).
Check: the track begins at \(g(f(x_0))=(1,0)\) and ends at \(x_0=(1,0)\). This confirms that in this choice of base point, the adjustment \(\alpha_\#\) really is the identity.
Worked Example: A Rotation with Different Chosen Base Points
Give \(S^1\) its usual subspace topology, choose \(u\in S^1\), and set \(f(z)=uz\) and \(g(w)=u^{-1}w\). Choose \(x_0=1\), so \(y_0=u\). Direct substitution gives \(g(f(z))=u^{-1}uz=z\) and \(f(g(w))=uu^{-1}w=w\); thus both composites are identities.
Here \(f\) sends a loop \(\gamma\) based at \(1\) to the loop \(t\mapsto u\gamma(t)\) based at \(u\). The map \(g\) sends that loop back to \(t\mapsto u^{-1}u\gamma(t)=\gamma(t)\), based at \(1\). In the reverse order, \(g\) first sends a loop based at \(u\) to one based at \(1\), and \(f\) returns it exactly to the original loop based at \(u\). Therefore the induced homomorphisms are inverse on loop classes, even though their base points differ.
Check: \(g(y_0)=u^{-1}u=1=x_0\), so the track for \(g\circ f\simeq\operatorname{id}\) may be constant. Also \(f(g(y_0))=u u^{-1}u=u=y_0\), so the other track may be constant. The formula for each composite agrees with the corresponding identity map.
Worked Example: Different Tracks for Identity Maps
Again let \(X=Y=S^1\) with the usual topology, take \(f=g=\operatorname{id}_{S^1}\), and choose \(x_0=y_0=1\). For the homotopy \(H(z,s)=e^{2\pi i s}z\), the map at \(s=0\) and the map at \(s=1\) are both the identity. Its track at \(1\) is the loop \(\alpha(s)=e^{2\pi i s}\). Choose the constant homotopy for \(K\), whose track \(\beta\) at \(1\) is constant.
The theorem’s proposed inverse is \(\Phi=\alpha_\#\circ\operatorname{id}_*\). For a loop class \([\gamma]\) at \(1\), its value is \([\bar\alpha*\gamma*\alpha]\). In other words, the moving track contributes conjugation by a loop; it cannot simply be discarded. The other composite is \(\alpha_\#\circ\beta_\#^{-1}=\alpha_\#\), an isomorphism because it is a base-point change. The proof shows that this composite is the identity, as it must be for the identity map’s inverse.
Check: both \(\alpha\) and \(\beta\) start and end at \(1\), so both base-point changes are maps from \(\pi_1(S^1,1)\) to itself. The track formula for \(K\) gives \(\beta_\#\circ(\operatorname{id})_*=\operatorname{id}\), consistent with \(\beta\) being constant.
A Common Base-Point Error
Common pitfall. It is not always meaningful to write \(g_*\circ f_*=\operatorname{id}\) without checking base points. For example, in the cylinder example choose \(x_0=(1,1)\). Then \(y_0=f(x_0)=1\), but \(g(y_0)=(1,0)\). Consequently \(f_*\) has target \(\pi_1(S^1,1)\), whereas \(g_*\) takes that group to \(\pi_1(S^1\times I,(1,0))\), not to \(\pi_1(S^1\times I,(1,1))\). Treating these as the same group and declaring the composite the identity is unjustified.
The correction is to use the track \(\alpha(s)=H((1,1),s)=(1,s)\) for the homotopy \(H((z,t),s)=(z,st)\). It runs from \(g(y_0)=(1,0)\) to \(x_0=(1,1)\). The correctly based inverse is \(\alpha_\#\circ g_*\). The homotopy track formula then gives \(\alpha_\#\circ g_*\circ f_*=\operatorname{id}\) on \(\pi_1(S^1\times I,(1,1))\).
The essential check in any application is therefore the endpoint of each path: identify the base point of the group before and after every induced map, then insert the homotopy track wherever those points differ.
Check Your Understanding
For each question, write the base points of the groups involved before composing induced maps.
- For a homotopy \(F_0\simeq F_1\), what is the direction of the base-point track, and which base-point change makes the two induced maps comparable?
- In the theorem, why does \(\alpha_\#\circ g_*\) have domain \(\pi_1(Y,y_0)\) and codomain \(\pi_1(X,x_0)\)?
- Explain why the homotopy \(K:f\circ g\simeq\operatorname{id}_Y\) implies that \((f\circ g)_*\) is an isomorphism, even when its track is not constant.
- In the cylinder example with \(x_0=(1,1)\), give the endpoints of the track \(\alpha\) and state which fundamental groups it connects.
- If both tracks in a homotopy inverse pair are constant at the chosen base points, what does the theorem say about the unadjusted induced maps \(f_*\) and \(g_*\)?