Why Fixing the Base Point Matters
A continuous map sends loops to loops only after its effect on the base point has been specified. If two maps agree on the base point and are connected by a homotopy that keeps it fixed at every time, then each loop and its two images fit together into a square whose two side edges are constant. That square is precisely a homotopy of loops fixing their endpoints. This observation proves that the induced homomorphisms on fundamental groups are equal.
Let \((X,x_0)\) and \((Y,y_0)\) be topological spaces with specified points. A loop based at \(x_0\) is a continuous map \(\alpha:I\to X\), where \(I=[0,1]\), such that \(\alpha(0)=\alpha(1)=x_0\). Two loops based at \(y_0\) represent the same element of \(\pi_1(Y,y_0)\) when they are homotopic through loops that keep both endpoints at \(y_0\). The fundamental group consists of these loop-homotopy classes, with multiplication given by concatenation.
For a based map \(f\), composing a loop \(\alpha\) based at \(x_0\) with \(f\) gives the loop \(f\circ\alpha\) based at \(y_0\). The induced homomorphism is defined by \(f_*([\alpha])=[f\circ\alpha]\), where brackets denote loop-homotopy classes. The well-definedness and homomorphism property of this construction are standard results from the earlier material on induced maps.
The Main Theorem
Proof. Let \(H:X\times I\to Y\) be a based homotopy from \(f\) to \(g\). To prove that the induced homomorphisms are equal, take an arbitrary element of \(\pi_1(X,x_0)\), represented by a loop \(\alpha:I\to X\) based at \(x_0\). Define $$ F:I\times I\to Y,\qquad F(s,t)=H(\alpha(s),t). $$ The map \(F\) is continuous because \((s,t)\mapsto(\alpha(s),t)\) and \(H\) are continuous. At the bottom and top edges, $$ F(s,0)=f(\alpha(s)),\qquad F(s,1)=g(\alpha(s)). $$ At the two side edges, the loop endpoints give $$ F(0,t)=H(x_0,t)=y_0,\qquad F(1,t)=H(x_0,t)=y_0. $$ Thus, for every \(t\), the path \(s\mapsto F(s,t)\) is a loop based at \(y_0\), and \(F\) is a homotopy of loops fixing the base point. It follows that \([f\circ\alpha]=[g\circ\alpha]\) in \(\pi_1(Y,y_0)\). Since \([\alpha]\) was arbitrary, \(f_*([\alpha])=g_*([\alpha])\) for every element of \(\pi_1(X,x_0)\), so \(f_*=g_*\). \(\square\)
The proof is a statement about every loop, not just a chosen representative of a special class. The base-point condition on \(H\) is what makes the side edges constant. Without it, the intermediate paths in the square may not be loops at \(y_0\), so they cannot be used directly as a homotopy in the definition of \(\pi_1(Y,y_0)\).
Worked Applications
Worked Example: A Perturbation of the Degree-Two Circle Map
Give the circle \(S^1=\{z\in\mathbb C:|z|=1\}\) its subspace topology from \(\mathbb C\), and base it at \(1\). Compare the maps \(f,g:S^1\to S^1\) given, using \(z=e^{2\pi i t}\), by $$ f(e^{2\pi i t})=e^{2\pi i(2t)},\qquad g(e^{2\pi i t})=e^{2\pi i(2t+\frac14\sin(2\pi t))}. $$
First check that the formula for \(g\) does not depend on the choice of \(t\). Replacing \(t\) by \(t+1\) changes \(2t+\frac14\sin(2\pi t)\) by \(2\), and multiplying the exponent by \(2\pi i\) makes no change to the complex value. Both maps send \(1\) to \(1\), using \(t=0\).
Define $$ H(e^{2\pi i t},s)=e^{2\pi i(2t+\frac{s}{4}\sin(2\pi t))},\qquad 0\leq s\leq1. $$ The same periodicity check applies for every \(s\), so \(H\) is well-defined on \(S^1\times I\). Its formulas at \(s=0\) and \(s=1\) are \(f\) and \(g\), respectively. At the base point \(t=0\), the sine term is \(0\), so \(H(1,s)=1\) for every \(s\). This is a based homotopy, and the theorem gives \(f_*=g_*\).
For an explicit point on the homotopy, take \(t=\frac14\). Then \(2t=\frac12\) and \(\sin(2\pi t)=\sin(\frac{\pi}{2})=1\), so $$ H(e^{\pi i/2},s)=e^{2\pi i(\frac12+\frac{s}{4})}. $$ At \(s=0\) this is \(e^{\pi i}=-1\); at \(s=1\) it is \(e^{3\pi i/2}=-i\). The endpoint values agree with \(f(i)=-1\) and \(g(i)=-i\), confirming the calculation.
In the standard identification \(\pi_1(S^1,1)\cong\mathbb Z\), the loop \(\ell(t)=e^{2\pi i t}\) represents the generator. Its image under \(f\) is \(e^{4\pi i t}\), which represents twice that generator. Since \(f_*=g_*\), \(g\) has the same action on this generator.
Worked Example: Two Different Maps into the Plane
Give \(\mathbb R^2\) its Euclidean topology and base it at \((0,0)\). Give \(S^1\) the topology and base point from the previous example. Consider $$ f(z)=z-1,\qquad g(z)=2(z-1), $$ where complex numbers are regarded as points of \(\mathbb R^2\). Both maps send \(1\) to \((0,0)\). The homotopy $$ H(z,s)=(1+s)(z-1) $$ joins them: \(H(z,0)=f(z)\) and \(H(z,1)=g(z)\). Also \(H(1,s)=0\) for all \(s\), so this homotopy is based. Therefore \(f_*=g_*\).
For a direct check on the loop \(\ell(t)=e^{2\pi i t}\), its images are $$ (f\circ\ell)(t)=(\cos(2\pi t)-1,\sin(2\pi t)),\qquad (g\circ\ell)(t)=(2\cos(2\pi t)-2,2\sin(2\pi t)). $$ The homotopy of these loops is $$ F(t,s)=((1+s)(\cos(2\pi t)-1),(1+s)\sin(2\pi t)). $$ At \(t=0\) and \(t=1\), \(\cos(2\pi t)=1\) and \(\sin(2\pi t)=0\), so both endpoints remain \((0,0)\) for every \(s\). At \(s=0\) and \(s=1\), the displayed loop formulas are exactly \(f\circ\ell\) and \(g\circ\ell\). Thus the loop classes agree, as the theorem predicts. The target \(\mathbb R^2\) is convex, and every loop based at the origin contracts by multiplying its values by \(1-s\); hence its fundamental group is trivial. The example also checks the conclusion in that special case.
Worked Example: A Based Homotopy on the Interval
Give \(I=[0,1]\) its subspace topology from \(\mathbb R\), and use \(0\) as base point in both source and target. Define \(f,g:I\to I\) by \(f(t)=t^2\) and \(g(t)=t\). Set $$ H(t,s)=(1-s)t^2+st. $$ At \(s=0\), this equals \(t^2\), and at \(s=1\), it equals \(t\). Also \(H(0,s)=0\). For \(t,s\in[0,1]\), both \(t^2\) and \(t\) lie in \([0,1]\), and \(H(t,s)\) is their convex combination, so \(H(t,s)\in[0,1]\). Thus \(H\) is a based homotopy and \(f_*=g_*\).
For the particular loop \(\alpha:I\to I\) based at \(0\) given by \(\alpha(r)=0\), both image loops are constant at \(0\). More generally, every loop \(\beta\) in \(I\) based at \(0\) contracts through based loops by \(K(r,u)=(1-u)\beta(r)\): its values stay in \([0,1]\), and \(K(0,u)=K(1,u)=0\). Consequently \(\pi_1(I,0)\) is trivial, which independently confirms that the two induced homomorphisms agree.
A Common Pitfall: The Base Point Can Move
Suppose \(S^1\) has its usual topology and base point \(1\), and consider the homotopy $$ H(z,s)=e^{2\pi i s}z. $$ At \(s=0\) and \(s=1\), the map is the identity map of \(S^1\). It is tempting to apply the main theorem to this homotopy. But its base point track is $$ H(1,s)=e^{2\pi i s}, $$ which is not constantly \(1\). For example, at \(s=\frac12\), it equals \(e^{\pi i}=-1\). The intermediate maps therefore send the base point to \(-1\) at that time, rather than to \(1\).
For the loop \(\ell(t)=e^{2\pi i t}\), the paths obtained at time \(s\) are \(t\mapsto e^{2\pi i s}e^{2\pi i t}\). Their endpoints equal \(e^{2\pi i s}\), not \(1\) for general \(s\). So these paths are not loops based at \(1\), and this homotopy does not provide a homotopy through based loops. The two endpoint maps in this particular example happen to be identical, but the argument using this \(H\) is not a proof of the theorem’s conclusion. The fixed-base-point hypothesis is not optional: it is exactly what makes the square in the proof have constant side edges.
Check Your Understanding
Use the definition of a based homotopy and the square in the proof to answer these questions.
- In the main theorem, what are the four boundary paths of \(F(s,t)=H(\alpha(s),t)\)?
- For \(S^1\) based at \(1\), verify that \(H(e^{2\pi i t},s)=e^{2\pi i(3t+\frac15s\sin(2\pi t))}\) is well-defined and fixes the base point.
- Why does a homotopy of maps that moves \(x_0\) fail to give, in general, a homotopy through loops based at one fixed point?
- Give a based homotopy from \(t\mapsto t^3\) to \(t\mapsto t\) on \(I\), based at \(0\), and verify that its values remain in \(I\).
- State precisely what equality \(f_*=g_*\) says about the images of an arbitrary loop class in \(\pi_1(X,x_0)\).