Homotopy Invariance of the Fundamental Group

What Homotopy Invariance Says

A continuous map sends loops to loops and therefore induces a homomorphism of fundamental groups. When two maps are joined by a homotopy, it is tempting to say that their induced homomorphisms are equal. That is exactly right when the homotopy keeps the chosen base point fixed. If the base point moves, however, the two image loops belong to fundamental groups based at different points. The homotopy still gives a precise comparison: transport one group to the other along the path traced by the base point.

We use the convention that a concatenation $\lambda*\mu$ traverses $\lambda$ first and $\mu$ second. The reverse path is written $\bar\lambda$. Paths are considered up to homotopy fixing their endpoints, and $[\lambda]$ denotes the resulting path-homotopy class. For a based map $f:(X,x_0)\to(Y,y_0)$, the induced homomorphism is $f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)$, given by $f_*([\gamma])=[f\circ\gamma]$.

Definition: A homotopy from $f:X\to Y$ to $g:X\to Y$ is a continuous map $H:X\times I\to Y$ such that $H(x,0)=f(x)$ and $H(x,1)=g(x)$ for every $x\in X$. Given a chosen point $x_0\in X$, its base-point track is the path $\alpha:I\to Y$ defined by $\alpha(t)=H(x_0,t)$. It runs from $y_0=f(x_0)$ to $y_1=g(x_0)$.

For a path $\alpha$ from $y_0$ to $y_1$, change of base point transports a loop at $y_0$ to a loop at $y_1$ by going backwards along $\alpha$, traversing the loop, and returning along $\alpha$. We write this map as $c_\alpha$.

Definition: The change-of-base-point map along $\alpha$ is $$ c_\alpha:\pi_1(Y,y_0)\longrightarrow\pi_1(Y,y_1), \qquad c_\alpha([\delta])=[\bar\alpha*\delta*\alpha]. $$ The displayed path starts and ends at $y_1$. This map is an isomorphism; its inverse is $c_{\bar\alpha}$, since the consecutive paths $\alpha*\bar\alpha$ and $\bar\alpha*\alpha$ contract, with endpoints fixed, by retracing $\alpha$.

The Path Relation from the Homotopy Square

The key fact is obtained by restricting $H$ to the image of a loop and looking at the resulting square. Its two routes from one corner to the opposite corner are homotopic while their endpoints stay fixed.

Lemma: Let $\gamma:I\to X$ be a loop based at $x_0$, and let $H$ be a homotopy from $f$ to $g$, with base-point track $\alpha$. Then the paths $f\circ\gamma*\alpha$ and $\alpha*g\circ\gamma$ are homotopic fixing endpoints.

Proof. Define $K:I\times I\to Y$ by $K(s,t)=H(\gamma(s),t)$. Both routes in the lemma begin at $y_0$ and end at $y_1$. To give an explicit homotopy between them, define paths $p,q:I\to I\times I$ by $$ p(u)= \begin{cases} (2u,0),&0\le u\le \tfrac12,\\ (1,2u-1),&\tfrac12\le u\le1, \end{cases} \qquad q(u)= \begin{cases} (0,2u),&0\le u\le \tfrac12,\\ (2u-1,1),&\tfrac12\le u\le1. \end{cases} $$ The path $K\circ p$ traverses the bottom edge and then the right edge of the square. Since $\gamma(1)=x_0$, those edges give $f\circ\gamma$ followed by $\alpha$. The path $K\circ q$ traverses the left edge and then the top edge, giving $\alpha$ followed by $g\circ\gamma$. For $u,v\in I$, let $R(u,v)=(1-v)p(u)+v q(u)$, where the addition and scalar multiplication are coordinatewise. The square $I\times I$ is convex, so $R(u,v)\in I\times I$. Thus $K(R(u,v))$ is a homotopy between $K\circ p$ and $K\circ q$. Since $p(0)=q(0)=(0,0)$ and $p(1)=q(1)=(1,1)$, the endpoints remain fixed throughout. $\square$

Prepend $\bar\alpha$ to both paths in this lemma. The resulting paths are still homotopic fixing endpoints. Associativity of path concatenation up to endpoint-fixed reparameterization gives $\bar\alpha*(f\circ\gamma*\alpha)\simeq(\bar\alpha*\alpha)*g\circ\gamma$. The path $\bar\alpha*\alpha$ goes from $y_1$ to $y_0$ and back to $y_1$, so it contracts to the constant path at $y_1$ by retracing $\alpha$. Consequently, $$ \bar\alpha*(f\circ\gamma)*\alpha\simeq g\circ\gamma, $$ with endpoints fixed. In particular, the direction of the change-of-base-point path is determined by the endpoints: $\bar\alpha$ must come first, because the output loop is based at $y_1$.

Homotopy Invariance with a Moving Base Point

Theorem: Let $H:X\times I\to Y$ be a homotopy from $f:X\to Y$ to $g:X\to Y$, and choose $x_0\in X$. Write $y_0=f(x_0)$, $y_1=g(x_0)$, and let $\alpha(t)=H(x_0,t)$. Then $$ g_*=c_\alpha\circ f_*: \pi_1(X,x_0)\longrightarrow\pi_1(Y,y_1). $$ Thus the two induced homomorphisms agree after the change-of-base-point isomorphism along the track.

Proof. Let $[\gamma]\in\pi_1(X,x_0)$, represented by a loop $\gamma$ at $x_0$. The lemma and the cancellation of $\bar\alpha*\alpha$ give $[g\circ\gamma]=[\bar\alpha*(f\circ\gamma)*\alpha]$. By the definitions of the induced homomorphism and change of base point, the left side is $g_*([\gamma])$, while the right side is $c_\alpha(f_*([\gamma]))$. This holds for every element $[\gamma]$ of $\pi_1(X,x_0)$, so the homomorphisms are equal. $\square$

The based case follows without an additional argument. If $H$ fixes the base point, then $\alpha$ is the constant path at $y_0=y_1$. Inserting or deleting constant paths does not change a path-homotopy class, so $c_\alpha$ is the identity homomorphism. Therefore $f_*=g_*$. If the track is not constant, the theorem gives an equality only after transporting between the two target fundamental groups; those groups have different specified base points.

Worked Example: A Homotopy That Fixes the Base Point

Let $X=Y=S^1=\{z\in\mathbb C:|z|=1\}$, with base point $1$, and set $H(z,t)=e^{2\pi i t}z$. At $t=0$, $f(z)=z$; at $t=1$, $g(z)=z$, because $e^{2\pi i}=1$. The track is $\alpha(t)=H(1,t)=e^{2\pi i t}$, which begins and ends at $1$, but is not a constant path.

For any loop $\gamma$ based at $1$, the theorem gives $[g\circ\gamma]=[\bar\alpha*(f\circ\gamma)*\alpha]$. Here $g=f$, so this states that conjugating the loop class $f_*([\gamma])$ by the track loop $\alpha$ leaves that class unchanged. This conclusion follows from the existence of the displayed homotopy, not from an assumption that every conjugation in every fundamental group is trivial. In this example, $f_*=g_*$ also follows directly because $f$ and $g$ are the same map.

Check: at the chosen base point, $f(1)=1$ and $g(1)=e^{2\pi i}=1$. Thus both induced maps have codomain $\pi_1(S^1,1)$, and the track is indeed a loop at $1$.

Worked Example: Rotating the Circle and Moving the Base Point

Again take $X=Y=S^1$ and $x_0=1$, but now define $H(z,t)=e^{2\pi i t/3}z$. Then $f(z)=z$ and $g(z)=\omega z$, where $\omega=e^{2\pi i/3}=\cos(2\pi/3)+i\sin(2\pi/3)=-\tfrac12+i\tfrac{\sqrt3}{2}$. The base-point track is $\alpha(t)=e^{2\pi i t/3}$, from $1$ to $\omega$.

For every loop $\gamma$ at $1$, the theorem yields the fully based comparison $[g\circ\gamma]=[\bar\alpha*(f\circ\gamma)*\alpha]\in\pi_1(S^1,\omega)$. The path on the right starts at $\omega$, follows $\bar\alpha$ from $\omega$ to $1$, traverses $f\circ\gamma$, and returns to $\omega$ along $\alpha$. The left side also starts and ends at $\omega$, since $g(1)=\omega$. This verifies both the orientation of the track and the base point of the resulting class.

Check: $H(z,0)=e^0z=z=f(z)$, while $H(z,1)=e^{2\pi i/3}z=\omega z=g(z)$. Also $|\omega|^2=(-1/2)^2+(\sqrt3/2)^2=1/4+3/4=1$, so the endpoint lies on $S^1$.

Worked Example: Homotopic Constant Maps

Let $X=S^1$, let $Y=\mathbb R$ with its standard topology, and choose $y_0=2$ and $y_1=5$. Define $f(z)=2$, $g(z)=5$, and $H(z,t)=2+3t$. This is a homotopy because $H(z,0)=2=f(z)$ and $H(z,1)=5=g(z)$, for every $z$. At $x_0=1$, the track is $\alpha(t)=2+3t$, from $2$ to $5$.

For a loop $\gamma$ at $1$, $f\circ\gamma$ is the constant path at $2$, and $g\circ\gamma$ is the constant path at $5$. The theorem says that the constant loop at $5$ is homotopic, with endpoints fixed, to $\bar\alpha$ followed by the constant loop at $2$ and then $\alpha$. The latter path traverses the interval from $5$ to $2$ and back to $5$, so it contracts by retracing. Thus both induced homomorphisms send every loop class to the identity element of the appropriate target group, and the change-of-base-point map sends the identity to the identity.

Check: $\alpha(0)=2+3(0)=2$ and $\alpha(1)=2+3(1)=5$. Therefore $\bar\alpha$ starts at $5$, and the transported loop begins and ends at $5$, as required.

Why the Conjugation Cannot Be Omitted

A common mistake is to conclude that every unbased homotopy gives $f_*=g_*$ as homomorphisms. When $f(x_0)\ne g(x_0)$, this equality is not even an equality between maps with the same target group: one homomorphism lands in $\pi_1(Y,f(x_0))$, the other in $\pi_1(Y,g(x_0))$. The rotation example makes this concrete: its two target groups are based at $1$ and at $-\tfrac12+i\tfrac{\sqrt3}{2}$, respectively. The correct comparison is $g_*=c_\alpha\circ f_*$.

Even when the track begins and ends at the same point, replacing the formula by $g_*=f_*$ can be wrong. In a figure-eight space, the fundamental group at the crossing is the free group on the two circle loops $a,b$. Conjugation by the loop $b$ sends $a$ to $b^{-1}ab$, which differs from $a$: these are distinct reduced words. Thus a base-point track that is a loop can produce a genuinely nontrivial conjugation. The theorem does not claim the two homomorphisms are equal in that situation; it identifies exactly how the track relates them.

Takeaway: A homotopy fixed at the chosen base point gives equal induced homomorphisms. For a homotopy whose base point moves along $\alpha$, first apply $f_*$, then change base point by $c_\alpha([\delta])=[\bar\alpha*\delta*\alpha]$; the result is $g_*$.

Check Your Understanding

For each question, keep track of the starting and ending points of every path.

For a homotopy $H$ and loop $\gamma$ at $x_0$, what are the endpoints of $f\circ\gamma*\alpha$ and $\alpha*g\circ\gamma$?
Write the base-point transport formula when $\alpha$ runs from $y_0$ to $y_1$. At which point is the transported loop based?
If $H(x_0,t)=y_0$ for every $t$, what does the moving-base-point theorem imply about $f_*$ and $g_*$?
In the rotation example, compute $g(1)$ and state the base point of $g\circ\gamma$.
Why does the unbased-homotopy formula not generally assert $f_*=g_*$ when the track starts and ends at the same point?

Homotopy Invariance of the Fundamental Group

What You'll Learn