Why a Homeomorphism Preserves the Fundamental Group
The fundamental group records based loops up to homotopy, together with the operation of concatenating loops. A homeomorphism preserves more than the points of a space: it and its inverse are continuous, so they carry loops and homotopies in both directions. This gives a precise reason that the fundamental groups at corresponding base points are isomorphic.
Let \(I=[0,1]\) with its standard topology. A loop based at \(x_0\) in a topological space \(X\) is a continuous map \(\gamma:I\to X\) with \(\gamma(0)=\gamma(1)=x_0\). Two such loops \(\gamma,\delta\) are based-homotopic if there is a continuous map \(H:I\times I\to X\) such that \(H(0,t)=\gamma(t)\), \(H(1,t)=\delta(t)\), and \(H(s,0)=H(s,1)=x_0\) for all \(s,t\in I\). The equivalence class of \(\gamma\) is written \([\gamma]\).
For a continuous map \(f:X\to Y\) with \(f(x_0)=y_0\), composition sends a loop \(\gamma\) based at \(x_0\) to the loop \(f\circ\gamma\) based at \(y_0\). It also sends a based homotopy \(H\) to the based homotopy \(f\circ H\). Thus the induced map is well-defined by \(\ f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)\), \(f_*([\gamma])=[f\circ\gamma]\). The based-map condition \(f(x_0)=y_0\) is essential: it ensures that the image loop has the required base point.
The Composition Law for Induced Maps
Proof. For any class \([\gamma]\in\pi_1(X,x_0)\), the definition gives \[ (g\circ f)_*([\gamma])=[(g\circ f)\circ\gamma]. \] Associativity of function composition gives \((g\circ f)\circ\gamma=g\circ(f\circ\gamma)\), so \[ [(g\circ f)\circ\gamma]=g_*([f\circ\gamma])=(g_*\circ f_*)([\gamma]). \] This proves the composition identity on every class. For the identity map, \(\operatorname{id}_X\circ\gamma=\gamma\), so \((\operatorname{id}_X)_*([\gamma])=[\gamma]\) for every class. \(\square\)
For completeness, induced maps preserve the group operation as well. Applying \(f\) to a concatenated loop gives the concatenation of the two image loops: on each half of \(I\), both maps are given by composing the corresponding half of the original loop with \(f\). Therefore \(f_*([\gamma][\delta])=f_*([\gamma])f_*([\delta])\). The preservation of based homotopies ensures this statement concerns classes, not just chosen loop representatives.
The Homeomorphism Invariance Theorem
Proof. Since \(h\) is continuous and \(h(x_0)=y_0\), it induces a homomorphism \(h_*\) with the stated domain and codomain. The inverse homeomorphism \(h^{-1}:Y\to X\) is continuous and satisfies \(h^{-1}(y_0)=x_0\), so it induces \((h^{-1})_*:\pi_1(Y,y_0)\to\pi_1(X,x_0)\). By the composition law, \[ (h^{-1})_*\circ h_*=(h^{-1}\circ h)_*=(\operatorname{id}_X)_*, \qquad h_*\circ(h^{-1})_*=(h\circ h^{-1})_*=(\operatorname{id}_Y)_*. \] Both identity-induced maps are identity maps on their respective fundamental groups. Hence the two induced homomorphisms are mutually inverse, which proves that \(h_*\) is an isomorphism. \(\square\)
The theorem concerns the groups at corresponding base points: the target base point must be \(h(x_0)\). It says that the groups are isomorphic, not that their loops are literally the same maps or that the groups are equal as sets. The inverse homeomorphism supplies the inverse correspondence, which is why continuity of a bijection alone would not suffice for this proof.
Worked Consequences
Worked Example: An Affine Homeomorphism of Intervals
Give \(X=[0,1]\) and \(Y=[2,5]\) their subspace topologies from \(\mathbb R\). Consider \(h:X\to Y\), \(h(x)=2+3x\), with base point \(x_0=1/3\).
The map has inverse \(h^{-1}(y)=(y-2)/3\). Both are continuous as restrictions of continuous affine maps on \(\mathbb R\), and for \(x\in[0,1]\) we have \(2\le 2+3x\le5\). Thus \(h\) is a homeomorphism. The corresponding base point is \[ h(1/3)=2+3(1/3)=3. \] The theorem gives an isomorphism \(h_*:\pi_1([0,1],1/3)\to\pi_1([2,5],3)\).
For a concrete loop, let \(\gamma(t)=1/3+(1/2)t(1-t)\). Its endpoints are \(1/3\). Since \(0\le t(1-t)\le1/4\) on \(I\), its values satisfy \(1/3\le\gamma(t)\le1/3+1/8=11/24\), so it lies in \([0,1]\). Its image loop is \[ (h\circ\gamma)(t)=2+3\left(1/3+(1/2)t(1-t)\right)=3+(3/2)t(1-t). \] Its endpoints are \(3\), and its maximum is \(3+(3/2)(1/4)=27/8\), which lies in \([2,5]\). Applying the inverse map recovers the original loop: \[ h^{-1}((h\circ\gamma)(t))=\frac{3+(3/2)t(1-t)-2}{3} =\frac13+\frac12t(1-t)=\gamma(t). \]
In fact, both fundamental groups here are trivial, as for any interval. For this particular loop, a based null-homotopy in \(X\) is \(H(s,t)=(1-s)\gamma(t)+s/3\). Its endpoints in \(t\) are always \(1/3\), and each value lies between \(1/3\) and \(11/24\), hence in \(X\). At \(s=0\) it is \(\gamma\); at \(s=1\) it is constant. The image loop is null-homotopic in \(Y\) by applying \(h\) to this homotopy. The inverse formula above checks that the induced correspondence returns each loop class to its original class.
Worked Example: Rotating the Circle
Give \(S^1=\{z\in\mathbb C:|z|=1\}\) its subspace topology from \(\mathbb C\). Define \(r:S^1\to S^1\) by \(r(z)=iz\). Multiplication by \(i\) preserves absolute value, since \(|iz|=|i||z|=|z|=1\), so the map takes \(S^1\) into itself. It is continuous, and its inverse \(r^{-1}(z)=-iz\) is continuous. Thus \(r\) is a homeomorphism.
At base point \(1\), the corresponding target base point is \(i\), since \(r(1)=i\). The theorem therefore gives \(r_*:\pi_1(S^1,1)\to\pi_1(S^1,i)\) as an isomorphism. For the loop \(\gamma(t)=e^{2\pi it}\), the endpoints are \(e^0=1\) and \(e^{2\pi i}=1\). Its image is \((r\circ\gamma)(t)=ie^{2\pi it}\), whose endpoints are \(i\) and \(i\). Applying the inverse map gives \[ r^{-1}((r\circ\gamma)(t))=(-i)(ie^{2\pi it})=e^{2\pi it}=\gamma(t). \] Thus this loop class is carried to a based loop class at \(i\) and recovered exactly by the inverse induced map.
The base point really changes under this rotation. Although this example does not calculate the structure of either fundamental group, it identifies the isomorphism between the groups at these two specific base points.
Worked Example: A Homeomorphism of Two-Point Discrete Spaces
Let \(X=\{a,b\}\) and \(Y=\{u,v\}\), each with the discrete topology, in which every subset is open. Define \(h(a)=u\) and \(h(b)=v\). The map and its inverse are continuous because every subset of each space is open. At base point \(a\), it induces an isomorphism \(\pi_1(X,a)\to\pi_1(Y,u)\).
Every continuous map from \(I\) to a discrete space is constant. Indeed, if a map \(\gamma:I\to X\) took both values, then \(\gamma^{-1}(\{a\})\) and \(\gamma^{-1}(\{b\})\) would be nonempty, disjoint open sets whose union is \(I\). This contradicts the connectedness of the interval \(I\). Thus the only loop based at \(a\) is the constant loop, and the only loop based at \(u\) is also constant. Both groups have one element, and \(h_*\) sends that element to that element. Applying \(h^{-1}\) returns it, verifying the isomorphism directly.
Common Pitfall: Forgetting the Base Point
A homeomorphism does not automatically give an induced map between fundamental groups with any base points one chooses. For example, for the rotation \(r(z)=iz\) on \(S^1\), take the source base point to be \(1\) and incorrectly try to define a map from \(\pi_1(S^1,1)\) to \(\pi_1(S^1,1)\) by \([\gamma]\mapsto[r\circ\gamma]\). The loop \(\gamma(t)=1\) is based at \(1\), but \(r\circ\gamma\) is the constant loop at \(i\), not a loop based at \(1\). So the proposed output is not an element of the claimed target group.
The correction is to use the corresponding base point \(r(1)=i\). Then \(r\circ\gamma\) is a loop based at \(i\), and the inverse rotation \(r^{-1}(z)=-iz\) gives the inverse induced map. The theorem requires both the correct base point and a homeomorphism; its proof uses the inverse map to establish that the induced map is invertible.
Check Your Understanding
Use the definition of an induced map and the composition law to answer the following.
- If \(h:X\to Y\) is a homeomorphism and \(x_0\in X\), what must the target base point be for \(h_*\) to be defined?
- Write the two compositions of induced maps that verify \((h^{-1})_*\) is the inverse of \(h_*\).
- For \(h(x)=4-2x\) from \([0,1]\) to \([2,4]\), find the image of the base point \(x_0=1/2\) and write the inverse function.
- Why does a continuous bijection not by itself provide the inverse induced map used in the theorem?
- In the circle rotation example, why does applying the rotation to a loop based at \(1\) produce a loop based at \(i\), rather than at \(1\)?