Why Well-Definedness Needs a Proof
The fundamental group is made from equivalence classes of based loops, not from individual loops. A continuous map can be applied to each loop, but to get a map on classes we must know that equivalent loops remain equivalent after applying it. The essential reason is that the same continuous map can be applied to every path in a homotopy, while preserving the fixed endpoints.
Use \(I=[0,1]\) with its standard topology, and give \(I\times I\) the product topology. Suppose \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) are topological spaces, with chosen points \(x_0\in X\) and \(y_0\in Y\). A based loop at \(x_0\) is a continuous map \(\alpha:I\to X\) with \(\alpha(0)=\alpha(1)=x_0\). A based homotopy from \(\alpha\) to another based loop \(\beta\) at \(x_0\) is a continuous map \(H:I\times I\to X\) such that
for every \(s,t\in I\). The first coordinate traces a loop, and the second coordinate changes that loop through the homotopy. The last two conditions require both endpoints to stay fixed at \(x_0\) throughout.
Earlier material defines \(\pi_1(X,x_0)\) as the set of based-homotopy classes of loops at \(x_0\). Write \([\alpha]\) for the class of \(\alpha\). The proposed induced map is \([f_*]([\alpha])=[f\circ\alpha]\); the conventional notation for this map is \(f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)\). This prescription is meaningful only if changing \(\alpha\) to an equivalent loop does not change the resulting class.
Composing a Based Homotopy
Proof. Since \(H\) and \(f\) are continuous, their composite \(K=f\circ H\) is continuous. In particular, for any open set \(V\subseteq Y\), \(K^{-1}(V)=H^{-1}(f^{-1}(V))\) is open: \(f^{-1}(V)\) is open in \(X\), and \(H\) is continuous. For every \(s\in I\), the boundary values in the homotopy parameter are \(K(s,0)=f(H(s,0))=f(\alpha(s))\) and \(K(s,1)=f(H(s,1))=f(\beta(s))\). For every \(t\in I\), the endpoint values are \(K(0,t)=f(H(0,t))=f(x_0)=y_0\) and \(K(1,t)=f(H(1,t))=f(x_0)=y_0\). Thus \(K\) is continuous, joins the two composite loops, and keeps both endpoints fixed at \(y_0\). \(\square\)
The proof uses both hypotheses on \(f\). Continuity ensures the composite is continuous; the condition \(f(x_0)=y_0\) ensures the endpoints land at the chosen point in \(Y\). No injectivity or surjectivity is needed.
Worked Example: A Nonlinear Map on the Plane
Give \(\mathbb{R}^2\) and \(\mathbb{R}\) their standard topologies. Let \(u=s(1-s)\) and define loops at \((0,0)\) by \(\alpha(s)=(s(1-s),0)\) and \(\beta(s)=(0,s(1-s))\). Define \(H:I\times I\to\mathbb{R}^2\) by \(H(s,t)=((1-t)u,tu)\). This is continuous because its coordinates are polynomials in \(s,t\). At \(s=0\) or \(s=1\), \(u=0\), so \(H\) equals \((0,0)\) at both endpoints for every \(t\). At \(t=0\), it equals \(\alpha(s)\); at \(t=1\), it equals \(\beta(s)\).
Now take \(f:\mathbb{R}^2\to\mathbb{R}^2\), \(f(x,y)=(x^2,y)\), which is continuous and sends \((0,0)\) to \((0,0)\). The composed homotopy is
At \(t=0\), this gives \((u^2,0)=f(\alpha(s))\); at \(t=1\), it gives \((0,u)=f(\beta(s))\). At \(s=0\) or \(s=1\), \(u=0\), so \(K(s,t)=(0,0)\) for every \(t\). As a point check, at \(s=1/2,\ t=1/2\), \(u=(1/2)(1-1/2)=1/4\), so \(K=( (1/2)^2(1/4)^2,(1/2)(1/4))=(1/64,1/8)\). Directly, \(H=(1/8,1/8)\) there and \(f(1/8,1/8)=((1/8)^2,1/8)=(1/64,1/8)\), as required.
Worked Example: Squaring a Circle-Valued Homotopy
Let \(S^1=\{z\in\mathbb{C}:|z|=1\}\) with the subspace topology from \(\mathbb{C}\), and use \(1\) as base point. Define \(\alpha(s)=e^{i(\pi/3)\sin(2\pi s)}\) and let \(\beta(s)=1\). These are loops based at \(1\), since \(\sin(0)=\sin(2\pi)=0\). The map \(H(s,t)=e^{i(1-t)(\pi/3)\sin(2\pi s)}\) is continuous and takes values in \(S^1\), since each value has modulus \(1\). It equals \(\alpha(s)\) at \(t=0\), equals \(\beta(s)\) at \(t=1\), and equals \(1\) when \(s=0\) or \(s=1\). Thus it is a based homotopy.
Take \(f:S^1\to S^1\) given by \(f(z)=z^2\). This map is continuous and \(f(1)=1\). Applying the theorem gives \(K(s,t)=e^{i(2\pi/3)(1-t)\sin(2\pi s)}\). At \(s=0\) and \(s=1\), the sine is zero and \(K=1\); at \(t=1\), \(K=1=f(\beta(s))\); and at \(t=0\), \(K=e^{i(2\pi/3)\sin(2\pi s)}=f(\alpha(s))\). For a numerical check at \(s=1/4,\ t=1/2\), \(\sin(2\pi s)=\sin(\pi/2)=1\), so \(K=e^{i\pi/3}=1/2+i\sqrt{3}/2\). Directly, \(H=e^{i\pi/6}=\sqrt{3}/2+i/2\); squaring gives \((\sqrt{3}/2+i/2)^2=1/2+i\sqrt{3}/2\), agreeing with the formula.
Why the Induced Map Is Well Defined
Proof. Let \([\alpha]=[\beta]\) in \(\pi_1(X,x_0)\). By the definition of these equivalence classes, there is a based homotopy \(H\) from \(\alpha\) to \(\beta\). The theorem gives a based homotopy from \(f\circ\alpha\) to \(f\circ\beta\), so \([f\circ\alpha]=[f\circ\beta]\). Therefore any two representatives of the same class produce the same class in \(Y\). This is exactly the condition that the rule on classes is well defined. \(\square\)
Worked Example: A Constant Based Map
Give \(\mathbb{R}\) its standard topology, take \(x_0=0\), and let \(Y=S^1\) with base point \(y_0=1\). Define \(f:\mathbb{R}\to S^1\) by \(f(x)=1\) for every \(x\); it is continuous and based. For illustration, let \(\alpha(s)=s(1-s)\), \(\beta(s)=0\), and \(H(s,t)=(1-t)s(1-s)\). These are loops at \(0\), and \(H\) is a based homotopy: it has the specified loops at \(t=0,1\), and \(H(0,t)=H(1,t)=0\).
For every \((s,t)\), \(K(s,t)=f(H(s,t))=1\). Thus both image loops are the constant loop at \(1\), and the composed homotopy stays at \(1\). For example, at \(s=1/2,\ t=1/2\), \(H=(1/2)(1/2)(1/2)=1/8\), but \(f(H)=1\). This confirms that a continuous based map need not preserve distinct loops as distinct loops; well-definedness only requires equivalent representatives to have equivalent images.
A Common Mistake: Letting the Base Point Move
The fixed-endpoint condition is not optional. Having based loops at the start and end of a proposed homotopy does not ensure that the intermediate paths remain loops at the same base point.
Common pitfall. In \(\mathbb{R}\) with its standard topology and base point \(0\), let \(\alpha(s)=\beta(s)=0\), and try \(H(s,t)=t(1-t)\). It does satisfy \(H(s,0)=\alpha(s)=0\) and \(H(s,1)=\beta(s)=0\), but \(H(0,t)=t(1-t)\), which is not always \(0\). For instance, at \(t=1/2\), \(H(0,1/2)=1/4\). So this is not a based homotopy. If \(f:\mathbb{R}\to\mathbb{R}\) is \(f(x)=2x\), then \(f(H(0,1/2))=1/2\), not the required base point \(0\). The error is to check only the top and bottom of the square and omit its two endpoint edges.
The valid construction checks all four boundary conditions: the two loop values at \(t=0,1\), and the two fixed endpoints for every \(t\). Once those conditions hold, continuity and \(f(x_0)=y_0\) ensure the corresponding conditions after composition. The algebraic group operation is not needed for this well-definedness argument; it concerns only the equivalence classes of loops.
Check Your Understanding
For each question, distinguish the homotopy parameter from the loop parameter and check both endpoint conditions.
- State the four boundary identities required for a based homotopy \(H\) from \(\alpha\) to \(\beta\).
- Why is the condition \(f(x_0)=y_0\) needed when composing a based homotopy with \(f\)?
- For \(f:\mathbb{R}\to\mathbb{R}\), \(f(x)=x^2\), and \(H(s,t)=(1-t)s(1-s)\), write down \(f(H(s,t))\) and check its values at \(t=0\), \(t=1\), \(s=0\), and \(s=1\).
- Does a continuous based map have to be injective for its induced map on classes to be well defined? Explain using the constant-map example.
- For \(H(s,t)=t(1-t)\), which based-homotopy condition fails, and what is its value at \(s=0,\ t=1/2\)?