Identity and Composition Laws for Induced Maps

Maps of Spaces and Maps of Loop Classes

A continuous map sends each loop in its domain to a loop in its codomain, provided we keep track of where the base point goes. The induced map on fundamental groups records this operation on homotopy classes. Two basic laws should hold: the identity map does nothing to a loop class, and applying two maps in succession has the same effect as applying their composite. The base points in these laws are part of the statement, not optional labels.

Let \(X\) and \(Y\) be topological spaces, and let \(x_0\in X\). A loop based at \(x_0\) is a continuous map \(\alpha:I\to X\), where \(I=[0,1]\) has its standard topology, such that \(\alpha(0)=\alpha(1)=x_0\). Two such loops are based homotopic if there is a continuous homotopy between them that keeps both endpoints at \(x_0\) throughout. The fundamental group \(\pi_1(X,x_0)\) consists of their based homotopy classes, written \([\alpha]\). Its product is \([\alpha][\beta]=[\alpha*\beta]\), where \(\alpha*\beta\) traverses \(\alpha\) on the first half of \(I\) and \(\beta\) on the second half.

The definition uses based homotopy classes, so it is important that the assignment does not depend on the representative loop. If \(H:I\times I\to X\) is a based homotopy from \(\alpha\) to \(\beta\), then \(f\circ H\) is a based homotopy from \(f\circ\alpha\) to \(f\circ\beta\): its endpoints remain \(f(x_0)=y_0\). Thus the assignment is well-defined. It is a homomorphism because composing a concatenated loop with \(f\) gives the concatenation of the two image loops, with the same half-interval parametrization.

The Identity Map

Proof. Let \([\alpha]\in\pi_1(X,x_0)\). By the definition of the induced map, \[ (\operatorname{id}_X)_*([\alpha]) =[\operatorname{id}_X\circ\alpha]. \] For every \(t\in I\), \((\operatorname{id}_X\circ\alpha)(t)=\alpha(t)\). Therefore the two loops are the same map and have the same based homotopy class, so \((\operatorname{id}_X)_*([\alpha])=[\alpha]\). This holds for every element of \(\pi_1(X,x_0)\), which is precisely the identity homomorphism. \(\square\)

This argument does not require the fundamental group to be nontrivial. If \(X\) has no nonconstant loops, the identity law still holds on its (possibly trivial) fundamental group. It also does not require any special property of \(X\): the identity map is continuous on every topological space and fixes every possible base point.

Composition of Based Maps

Suppose \(f:X\to Y\) and \(g:Y\to Z\) are continuous. To define their induced maps, choose \(x_0\in X\), put \(y_0=f(x_0)\), and put \(z_0=g(y_0)\). Then \(f\) is based from \((X,x_0)\) to \((Y,y_0)\), \(g\) is based from \((Y,y_0)\) to \((Z,z_0)\), and \(g\circ f\) is based from \((X,x_0)\) to \((Z,z_0)\).

Proof. Take any \([\alpha]\in\pi_1(X,x_0)\). Applying the definition first to the composite gives \[ (g\circ f)_*([\alpha])=[(g\circ f)\circ\alpha]. \] Applying the induced maps successively gives \[ (g_*\circ f_*)([\alpha]) =g_*( [f\circ\alpha]) =[g\circ(f\circ\alpha)]. \] For every \(t\in I\), the two representative loops have the same value: \[ ((g\circ f)\circ\alpha)(t) =g(f(\alpha(t))) =(g\circ(f\circ\alpha))(t). \] They are therefore the same loop, so their based homotopy classes agree. This proves equality on every element of \(\pi_1(X,x_0)\), and hence equality of the two homomorphisms. The base point of each resulting loop is \(g(f(x_0))=z_0\), as required for both sides to have the stated codomain. \(\square\)

The order matters: \(f_*\) acts first on a class in \(\pi_1(X,x_0)\), and then \(g_*\) acts on the resulting class in \(\pi_1(Y,y_0)\). This is the order dictated by the function composite \(g\circ f\), which applies \(f\) first and \(g\) second.

A Common Base-Point and Order Error

Common pitfall. The notation \(f_*\) does not define one homomorphism independently of base points. For example, let \(f:S^1\to S^1\) be \(f(z)=z^2\), but now choose \(i\) as the domain base point. Then \(f(i)=i^2=-1\). The constant loop at \(i\) is based at \(i\), and its image under \(f\) is the constant loop at \(-1\). It is an element of \(\pi_1(S^1,-1)\), not a loop based at \(1\). Writing this image as an element of \(\pi_1(S^1,1)\) is an error: the endpoint condition for a loop based at \(1\) is not satisfied.

There is a related order error. For \(f:X\to Y\) and \(g:Y\to Z\), the expression \(g_*\circ f_*\) is defined on \(\pi_1(X,x_0)\) and lands in \(\pi_1(Z,z_0)\). Reversing the order generally does not even make sense: \(g_*\) outputs a class in \(\pi_1(Z,z_0)\), whereas \(f_*\) expects an input in \(\pi_1(X,x_0)\). The correct order follows the successive action on loops: first \(f\), then \(g\).

Together, the two laws can be recorded as \[ (\operatorname{id}_X)_*=\operatorname{id}_{\pi_1(X,x_0)} \quad\text{and}\quad (g\circ f)_*=g_*\circ f_*, \] with all base points matched as specified. These are direct consequences of how maps act on parametrized loops. They provide a consistency check whenever a fundamental-group homomorphism is built by composing continuous maps.