Induced Maps on Fundamental Groups

From a Map of Spaces to a Map of Loop Classes

A continuous map transforms paths by composition: follow a path in the source, then apply the map to each point of that path. For loops, the base point matters. If the source loop starts and ends at $x_0$, its image starts and ends at $f(x_0)$. Thus the natural map is between fundamental groups based at these two points, which need not be the same.

Throughout, $I=[0,1]$ has its usual subspace topology from $\mathbb{R}$. A based loop at $x_0$ in a topological space $X$ is a continuous map $\gamma:I\to X$ such that $\gamma(0)=\gamma(1)=x_0$. Two such loops are based-homotopic if there is a continuous $H:I\times I\to X$ with $H(s,0)=\gamma(s)$, $H(s,1)=\eta(s)$, and $H(0,t)=H(1,t)=x_0$ for all $s,t\in I$. The fundamental group $\pi_1(X,x_0)$ is the set of based-homotopy classes $[\gamma]$ of based loops at $x_0$, with multiplication given by loop concatenation.

Definition: Let $f:X\to Y$ be continuous, and let $x_0\in X$. Set $y_0=f(x_0)$. The induced assignment is $$ f_*:\pi_1(X,x_0)\longrightarrow\pi_1(Y,y_0), \qquad f_*([\gamma])=[f\circ\gamma]. $$ Here $(f\circ\gamma)(t)=f(\gamma(t))$. The standard well-definedness theorem states that this assignment does not depend on the choice of representative $\gamma$ of its class; its proof is treated separately. Thus the formula defines a map on classes.

There are two checks built into this definition. First, $f\circ\gamma$ must be a loop at $y_0$. Second, its class must depend only on $[\gamma]$, not on the chosen loop representing that class. The first check follows directly from continuity and the endpoints. The second is the well-definedness issue just identified; calculations below use the induced map with that standard result understood.

Proposition: If $f:X\to Y$ is continuous and $\gamma$ is a based loop at $x_0$, then $f\circ\gamma$ is a based loop at $f(x_0)$.

Proof. The composition of continuous maps is continuous: for every open set $V\subseteq Y$, $(f\circ\gamma)^{-1}(V)=\gamma^{-1}(f^{-1}(V))$, which is open in $I$ because $f^{-1}(V)$ is open in $X$ and $\gamma$ is continuous. At the endpoints, $(f\circ\gamma)(0)=f(\gamma(0))=f(x_0)$ and $(f\circ\gamma)(1)=f(\gamma(1))=f(x_0)$. So $f\circ\gamma$ is a continuous loop based at $f(x_0)$. $\square$

Calculating Images of Circle Loops

Write $S^1=\{z\in\mathbb{C}:|z|=1\}$ with the subspace topology inherited from the usual topology on $\mathbb{C}$. For each integer $n$, define a loop based at $1$ by $\gamma_n(t)=e^{2\pi i n t}$, for $t\in I$. Its endpoints are both $1$, since $e^0=1$ and $e^{2\pi i n}=1$ for every integer $n$. The parametrization records how the loop is traversed; in particular, negative $n$ gives the reverse direction.

Worked Example: The Cubing Map

Let $f:S^1\to S^1$ be $f(z)=z^3$, with both source and target based at $1$. This map is continuous because it is the restriction to $S^1$ of a polynomial map on $\mathbb{C}$, and $f(1)=1$. Calculate $f_*([\gamma_2])$.

At each $t\in I$, composition gives $(f\circ\gamma_2)(t)=(e^{2\pi i(2)t})^3=e^{12\pi i t}$. Since $12\pi i t=2\pi i(6)t$, this is $\gamma_6(t)$. Therefore $f_*([\gamma_2])=[\gamma_6]$.

Check: at $t=0$, the image is $e^0=1=f(1)$; at $t=1$, it is $e^{12\pi i}=1=f(1)$. The calculated loop has the required target base point. The exponent $6$ also agrees with multiplying the input exponent $2$ by the power $3$.

Worked Example: Complex Conjugation

Let $c:S^1\to S^1$ be $c(z)=\overline{z}$. Coordinatewise, this is $(x,y)\mapsto(x,-y)$, so it is continuous, and $c(1)=1$. Calculate the image of $[\gamma_3]$.

For $t\in I$, $(c\circ\gamma_3)(t)=\overline{e^{2\pi i(3)t}}=e^{-6\pi i t}$. The last expression equals $\gamma_{-3}(t)$, so $c_*([\gamma_3])=[\gamma_{-3}]$.

Check: at $t=0$, $e^0=1$, and at $t=1$, $e^{-6\pi i}=1$, so both endpoints equal the target base point. The sign changes because conjugation sends $e^{i\theta}$ to $e^{-i\theta}$; it reverses the angular direction.

These calculations determine the image class by producing an explicit representative. They do not require identifying every element of $\pi_1(S^1,1)$ with an integer. In particular, the equation $f\circ\gamma_n=\gamma_{kn}$ for a power map is an equality of parametrized loops, and therefore directly gives the stated class.

The Target Base Point Need Not Be the Source Base Point

Worked Example: A Rotation with a Different Target Base Point

Define $r:S^1\to S^1$ by $r(z)=iz$. Multiplication by $i$ is a continuous rotation, and $r(1)=i$. Thus the induced assignment has source $\pi_1(S^1,1)$ and target $\pi_1(S^1,i)$, not $\pi_1(S^1,1)$. Calculate the image of $[\gamma_2]$.

For $t\in I$, $(r\circ\gamma_2)(t)=i e^{4\pi i t}=e^{i\pi/2}e^{4\pi i t}=e^{i(\pi/2+4\pi t)}$. At $t=0$, this is $e^{i\pi/2}=i$. At $t=1$, it is $e^{i(9\pi/2)}=e^{i\pi/2}=i$, because $9\pi/2=\pi/2+4\pi$. Hence it is a loop based at $i$, and $r_*([\gamma_2])=[\,t\mapsto i e^{4\pi i t}\,]\in\pi_1(S^1,i)$.

Check: writing the image as $e^{i(\pi/2+4\pi t)}$ gives the same endpoint at $t=1$ as at $t=0$, since the angles differ by $4\pi$, two full turns. The initial point is $i$, not $1$, exactly as the target group indicates.

In this example, the image loop is not one of the loops $\gamma_n$ as defined above, because those loops are based at $1$. It is nevertheless a perfectly valid representative of a class in $\pi_1(S^1,i)$. To compare it with loops based at $1$ requires an additional construction for changing base points; the present formula does not perform that change.

A Map from the Circle to the Torus

Let $T^2=S^1\times S^1$ with the product topology, and base it at $(1,1)$. A loop in $T^2$ is a pair of coordinate loops: its first coordinate is a loop in the first copy of $S^1$, and its second coordinate is a loop in the second copy. This makes maps into the torus convenient for explicit calculations.

Worked Example: Two Coordinate Winding Patterns

Define $F:S^1\to T^2$ by $F(z)=(z^2,\overline{z})$. Its coordinates are continuous, so $F$ is continuous for the product topology, and $F(1)=(1,1)$. Calculate $F_*([\gamma_3])$.

For $t\in I$, the first coordinate is $(e^{2\pi i(3)t})^2=e^{12\pi i t}$. The second coordinate is $\overline{e^{2\pi i(3)t}}=e^{-6\pi i t}$. Consequently, $(F\circ\gamma_3)(t)=(e^{12\pi i t},e^{-6\pi i t})$, and $F_*([\gamma_3])=[\,t\mapsto(e^{12\pi i t},e^{-6\pi i t})\,]$.

Check: at $t=0$, both coordinates are $1$, giving $(1,1)$. At $t=1$, they are $e^{12\pi i}=1$ and $e^{-6\pi i}=1$, again giving $(1,1)$. The first coordinate is $\gamma_6$, while the second is $\gamma_{-3}$; the two coordinates therefore have the expected exponents $2\cdot3=6$ and $-3$.

Concatenation at the Representative Level

One useful calculation is that applying a map to a concatenated loop gives exactly the concatenation of the image loops. For based loops $\gamma,\eta:I\to X$ at $x_0$, use the convention $(\gamma*\eta)(t)=\gamma(2t)$ for $0\le t\le 1/2$, and $(\gamma*\eta)(t)=\eta(2t-1)$ for $1/2\le t\le1$. Both formulas agree at $t=1/2$, because $\gamma(1)=x_0=\eta(0)$.

Proposition: If $f:X\to Y$ is continuous and $\gamma,\eta$ are based loops at $x_0$, then $f\circ(\gamma*\eta)=(f\circ\gamma)*(f\circ\eta)$ as parametrized loops at $f(x_0)$.

Proof. For $0\le t\le1/2$, the left side is $f((\gamma*\eta)(t))=f(\gamma(2t))$, which is the first piece of $((f\circ\gamma)*(f\circ\eta))(t)$. For $1/2\le t\le1$, the left side is $f(\eta(2t-1))$, which is its second piece. At the joining value $t=1/2$, both descriptions give $f(x_0)$. Thus the loops agree for every $t\in I$. $\square$

This is an equality of representatives, not merely a comparison of their classes. Once the induced assignment is regarded as a well-defined map on classes, the equality explains why it respects the concatenation operation. The calculation also identifies exactly where the base point enters: both image loops join at $f(x_0)$.

A Common Base-Point Mistake

A frequent error is to use the correct composition $f\circ\gamma$ but assign its class to the group based at the wrong point. For the rotation $r(z)=iz$, the incorrect calculation is $r_*([\gamma_2])=[\,t\mapsto i e^{4\pi i t}\,]\in\pi_1(S^1,1)$. This cannot be right as written: at $t=0$, the displayed loop has value $i$, while a loop based at $1$ must have value $1$ at $t=0$. At $t=1$ its value is also $i$, not $1$.

The correction is to track the base point before naming the target group. Since $r(1)=i$, the composition is based at $i$, and the correct statement is $r_*:\pi_1(S^1,1)\to\pi_1(S^1,i)$, with $r_*([\gamma_2])=[\,t\mapsto i e^{4\pi i t}\,]$. More generally, whenever a continuous map is applied to a based loop, compute $f(x_0)$ first. It determines where the image loop begins and ends, and therefore which fundamental group contains its class.

Check Your Understanding

In each calculation, state the target base point as well as the representative image loop.

For $p:S^1\to S^1$, $p(z)=z^4$, based at $1$, calculate $p_*([\gamma_{-2}])$.
For $c(z)=\overline{z}$, calculate the representative of $c_*([\gamma_{-1}])$.
For $r(z)=e^{i\pi/3}z$, identify the target fundamental group for the induced assignment from $\pi_1(S^1,1)$.
For $F(z)=(z,\overline{z})$ from $S^1$ to $T^2$, calculate $F\circ\gamma_2$ and check its endpoints.
Using the given piecewise definition of concatenation, verify the representative-level concatenation identity on each half of $I$.