Changing the Base Point
The fundamental group records loops based at a chosen point. If a space is path-connected, any two points can be joined by a path, so loops based at one point can be transported to loops based at the other. This gives an isomorphism between the two fundamental groups. The choice of path matters: different paths can produce different isomorphisms, even though the groups remain isomorphic.
Let \(X\) be a topological space. A path from \(x\) to \(y\) is a continuous map \(\gamma:I\to X\) with \(\gamma(0)=x\) and \(\gamma(1)=y\). A loop based at \(x\) is a path whose two endpoints are \(x\). Two loops based at \(x\) represent the same element of \(\pi_1(X,x)\) when they are homotopic while fixing their endpoints. We use the multiplication convention that \([\alpha][\beta]\) means first traverse \(\alpha\), then traverse \(\beta\). The reverse of a path \(\gamma\) is \(\gamma^{-1}(t)=\gamma(1-t)\). We use the standard loop-concatenation facts that multiplication is associative on homotopy classes and that a path followed by its reverse represents the identity.
Why the Base-Point Change Map Is an Isomorphism
Proof. First we check that the formula is well-defined. If \(\alpha\) and \(\beta\) are homotopic loops based at \(x_0\), the homotopy fixes their endpoints. Concatenating \(\gamma^{-1}\) before each loop and \(\gamma\) after it gives a homotopy of loops based at \(x_1\), still fixing the endpoints. Thus the resulting loop classes agree.
For loops \(\alpha,\beta\) based at \(x_0\), associativity of concatenation on classes and cancellation of \([\gamma][\gamma^{-1}]\) give $$ C_\gamma([\alpha][\beta]) =[\gamma^{-1}][\alpha][\beta][\gamma] =([\gamma^{-1}][\alpha][\gamma])([\gamma^{-1}][\beta][\gamma]). $$ So \(C_\gamma\) is a homomorphism. The reverse path \(\gamma^{-1}\) runs from \(x_1\) to \(x_0\). The map \(C_{\gamma^{-1}}\) is an inverse: for \([\alpha]\in\pi_1(X,x_0)\), $$ C_{\gamma^{-1}}(C_\gamma([\alpha])) =[\gamma][\gamma^{-1}][\alpha][\gamma][\gamma^{-1}] =[\alpha], $$ and the same cancellation gives \(C_\gamma(C_{\gamma^{-1}}([\beta]))=[\beta]\) for every \([\beta]\in\pi_1(X,x_1)\). Therefore \(C_\gamma\) is an isomorphism. \(\square\)
Worked Example: Changing Base Point in an Interval
Give \(I=[0,1]\) its subspace topology from \(\mathbb{R}\), and take \(x_0=0\), \(x_1=1\), and \(\gamma(t)=t\). The base-point change map is $$ C_\gamma([\alpha])=[\gamma^{-1}][\alpha][\gamma]. $$ Here \(\alpha\) is any loop based at \(0\), and the resulting loop is based at \(1\).
The interval is path-connected: for \(a,b\in I\), the path \(t\mapsto (1-t)a+tb\) stays in \(I\) and joins \(a\) to \(b\). Also, every loop in \(I\) contracts to its base point while fixing that point. For a loop \(\alpha\) based at \(c\in I\), the formula \(H(s,t)=(1-s)\alpha(t)+sc\) stays in \(I\), fixes \(t=0,1\), and deforms \(\alpha\) to the constant loop at \(c\). Thus both fundamental groups have only the identity element, and \(C_\gamma\) sends that identity to the identity.
There is no dependence on the path in this example. Indeed, if \(\eta\) is any other path from \(0\) to \(1\), the homotopy \(H(s,t)=(1-s)\gamma(t)+s\eta(t)\) stays in \(I\) and fixes both endpoints. The paths are homotopic with endpoints fixed, so \([\gamma^{-1}\eta]\) is the identity loop class at \(1\). As shown below, that makes the two base-point change maps equal.
Check: the displayed contraction has \(H(0,t)=\alpha(t)\), \(H(1,t)=c\), \(H(s,0)=c\), and \(H(s,1)=c\). These equalities verify the initial loop, final constant loop, and fixed endpoints.
How the Choice of Path Changes the Isomorphism
Proof. The loop representing \(h^{-1}\) is \(\eta^{-1}\gamma\). Write \(a=[\alpha]\). Using the definition of the maps and cancelling each path followed by its reverse gives $$ h^{-1}C_\gamma(a)h =[\eta^{-1}][\gamma][\gamma^{-1}][\alpha][\gamma][\gamma^{-1}][\eta] =[\eta^{-1}][\alpha][\eta] =C_\eta(a). $$ This proves the formula. Since \(C_\gamma\) is onto, its values range over all elements \(b\) of \(\pi_1(X,x_1)\). Therefore \(C_\eta=C_\gamma\) exactly when \(h^{-1}bh=b\) for every such \(b\). This condition is equivalent to \(h\) commuting with every \(b\), which is precisely that \(h\) lies in the center. \(\square\)
The theorem separates two statements that are easy to conflate. Path-connectedness guarantees an isomorphism between the groups at any two chosen points. It does not, by itself, select a preferred path or a preferred isomorphism. The groups are isomorphic, but the isomorphism supplied by this construction can depend on the path.
Worked Example: Different Paths on the Circle
Let \(X=S^1\subset\mathbb{R}^2\) have the subspace topology, and use the point \(x_0=x_1=(1,0)\) as both base points. The circle is path-connected: any two points can be joined by a circular arc. Let \(\gamma\) be the constant path and let \(\eta\) travel once around the circle in the positive direction. These paths have the same endpoints but are not homotopic with endpoints fixed.
The standard computation \(\pi_1(S^1,(1,0))\cong\mathbb{Z}\) identifies the once-around loop with a generator. This computation is a standard result whose proof is treated with the fundamental group calculations for the circle. Under this identification, the group is abelian, so every element is central. Since \(C_\gamma\) is the identity map, the path comparison formula gives \(C_\eta(a)=h^{-1}ah=a\) for every \(a\). Thus the paths differ, but the induced isomorphisms agree.
Check: the formula uses \(h=[\gamma^{-1}\eta]=[\eta]\), the once-around class. In an abelian group \(h^{-1}ah=a\), because \(ah=ha\) implies \(h^{-1}ah=a\).
Worked Example: A Path Choice That Changes the Map
For a standard example with a nonabelian fundamental group, let \(X=S^1\vee S^1\) be the wedge of two circles: take the disjoint union of two copies of \(S^1\), identify their points \((1,0)\), and give the result the quotient topology. Call the identified point \(x_0\). The standard computation by the Seifert–van Kampen theorem gives \(\pi_1(X,x_0)\) as the free group on the two circle loops \(a\) and \(b\); in particular, distinct reduced words represent distinct elements. This computation is cited here as a standard result; its proof belongs to the treatment of Seifert–van Kampen.
Take \(\gamma\) to be the constant path at \(x_0\), and take \(\eta\) to be the loop \(a\). Both are paths from \(x_0\) to itself. Then \(C_\gamma\) is the identity, while \(C_\eta(g)=a^{-1}ga\). For the element \(b\), the two outputs are $$ C_\gamma(b)=b,\qquad C_\eta(b)=a^{-1}ba. $$ The word \(a^{-1}ba\) is reduced: no adjacent letters are inverses. It differs from the one-letter reduced word \(b\). Hence these two isomorphisms are not equal.
Check: the path-comparison loop is \(h=[\gamma^{-1}\eta]=a\), so the formula gives \(C_\eta(b)=h^{-1}C_\gamma(b)h=a^{-1}ba\), as computed directly.
A Common Mistake
Common pitfall. Reversing the conjugation order gives the wrong path-comparison formula. With \(h=[\gamma^{-1}\eta]\), it is incorrect to claim \(C_\eta(a)=hC_\gamma(a)h^{-1}\). In the wedge-of-circles example, this would give \(aba^{-1}\) for \(C_\eta(b)\), whereas the defining formula gives \(a^{-1}ba\). Both words are reduced and they are different: the first begins with \(a\), the second with \(a^{-1}\). The correct order follows from \(h^{-1}=[\eta^{-1}\gamma]\), so \(h^{-1}C_\gamma(a)h=[\eta^{-1}\alpha\eta]=C_\eta(a)\).
Path-connectedness therefore gives base-point independence in the sense of abstract group isomorphism, not necessarily equality of groups with a canonical identification. A path supplies the identification; changing it conjugates the result, and the identification is unchanged precisely when the loop measuring the path difference is central.
Check Your Understanding
Use the path-concatenation convention stated above.
- For a path \(\gamma:x_0\to x_1\) and a loop \(\alpha\) based at \(x_0\), explain why \(\gamma^{-1}\alpha\gamma\) is based at \(x_1\).
- Write the inverse of \(C_\gamma\) and verify the cancellation that makes it an inverse.
- For paths \(\gamma,\eta:x_0\to x_1\), identify the loop \(h\) based at \(x_1\) that compares their base-point change maps.
- State the necessary and sufficient condition for \(C_\eta=C_\gamma\).
- Why does a path-connected space guarantee isomorphic fundamental groups at two points without specifying a unique isomorphism?