Why a Well-Defined Operation Needs a Proof
Concatenating two based loops means traversing the first loop and then the second. The fundamental group is formed from homotopy classes of based loops, rather than from individual loops. For concatenation to give an operation on those classes, its result must not depend on which representatives are chosen. The key point is that homotopies of the two loops must keep their endpoints fixed at the base point: this makes the two pieces of the concatenated homotopy meet continuously.
Let \((X,\mathcal{T})\) be a topological space and fix \(x_0\in X\). A path is a continuous map from \(I=[0,1]\) to \(X\). A based loop at \(x_0\) is a path \(\alpha:I\to X\) with \(\alpha(0)=\alpha(1)=x_0\). We use the usual concatenation, which allocates the first half of the parameter interval to the first loop and the second half to the second loop.
For based loops \(\alpha,\beta\) at \(x_0\), their concatenation is the loop \(\alpha*\beta:I\to X\) defined by
At the shared parameter value \(s=\frac12\), the first formula gives \(\alpha(1)=x_0\), while the second gives \(\beta(0)=x_0\). Thus the formulas agree there. Each formula is continuous on its half-interval, so the pasting argument below shows that \(\alpha*\beta\) is continuous. Its endpoints are \(\alpha(0)=x_0\) and \(\beta(1)=x_0\), so it is again a based loop.
The Piecewise Homotopy Principle
We will paste two homotopies across the vertical line \(s=\frac12\) in the square \(I\times I\). The elementary principle used here is the finite closed-set pasting lemma.
Proof. Let \(C\) be any closed subset of \(X\). For each \(i\), continuity of \(f_i\) makes \(f_i^{-1}(C)\) closed in \(D_i\). Since \(D_i\) is closed in \(Z\), a set closed in \(D_i\) is closed in \(Z\). The preimage \(f^{-1}(C)\) is the finite union of the sets \(f_i^{-1}(C)\), so it is closed in \(Z\). The closed-set criterion for continuity now gives that \(f\) is continuous. Agreement on overlaps ensures that \(f\) is a well-defined map. \(\square\)
In our application, the domains are \(D_1=[0,\frac12]\times I\) and \(D_2=[\frac12,1]\times I\). They are closed in the square, cover the square, and overlap exactly along \(s=\frac12\). Therefore it is enough to check that the two formulas agree at that seam.
Homotopies of Both Factors Give a Homotopy of the Concatenation
Proof. Let \(H:I\times I\to X\) be an endpoint-preserving homotopy from \(\alpha\) to \(\alpha'\), and let \(K:I\times I\to X\) be an endpoint-preserving homotopy from \(\beta\) to \(\beta'\). Define \(F:I\times I\to X\) by
Each formula is continuous on its closed half of the square, because it is a continuous homotopy composed with a continuous change of parameters. At the seam, the first formula gives \(H(1,t)=x_0\), and the second gives \(K(0,t)=x_0\). The formulas therefore agree. The finite closed-set pasting lemma shows that \(F\) is continuous.
For every \(t\), the endpoints are \(F(0,t)=H(0,t)=x_0\) and \(F(1,t)=K(1,t)=x_0\). At \(t=0\), its two pieces are \(\alpha(2s)\) and \(\beta(2s-1)\), so \(F(s,0)=(\alpha*\beta)(s)\). At \(t=1\), they are \(\alpha'(2s)\) and \(\beta'(2s-1)\), so \(F(s,1)=(\alpha'*\beta')(s)\). Thus \(F\) is an endpoint-preserving homotopy from \(\alpha*\beta\) to \(\alpha'*\beta'\), as required. \(\square\)
The theorem proves exactly the representative-independence needed for the operation on classes: replacing either loop by an endpoint-preserving homotopic loop leaves the class of the concatenation unchanged. It does not assert any further algebraic law for that operation.
Worked Example: Changing the First Loop in the Plane
Give \(\mathbb{R}^2\) its standard topology and take \(x_0=(0,0)\). Consider the based loops
All three have value \((0,0)\) at \(s=0\) and \(s=1\). Define \(H(s,t)=((1-t)s(1-s),\,t s(1-s))\). This is continuous, \(H(s,0)=\alpha(s)\), and \(H(s,1)=\alpha'(s)\). At either endpoint \(s=0\) or \(s=1\), the factor \(s(1-s)\) is zero, so \(H(0,t)=H(1,t)=(0,0)\).
Keep \(\beta\) fixed, using \(K(u,t)=\beta(u)\). The theorem gives the homotopy
Check the seam: at \(s=\frac12\), the first formula uses \(2s=1\), so its factor \(2s(1-2s)\) is zero. The second uses \(2s-1=0\), so its factor \((2s-1)(1-(2s-1))\) is also zero. Both give \((0,0)\). Check a point in each piece: at \(s=\frac14\), the first piece uses \(2s=\frac12\), so it gives \(((1-t)/4,t/4)\). At \(s=\frac34\), the second uses \(2s-1=\frac12\), so it gives \((0,2(1/2)(1/2))=(0,1/2)\). Finally, \(t=0\) gives \(\alpha*\beta\), and \(t=1\) gives \(\alpha'*\beta\). The endpoints remain \((0,0)\) throughout.
Worked Example: Changing Only the Second Loop
Again take \(X=\mathbb{R}^2\) with the standard topology and base point \((0,0)\). Let \(\alpha(s)=(s(1-s),0)\), \(\beta(s)=(0,s(1-s))\), and \(\beta'(s)=(0,-s(1-s))\). A homotopy from \(\beta\) to \(\beta'\) is \(K(u,t)=(0,(1-2t)u(1-u))\). Indeed, substituting \(t=0\) gives \(\beta\), substituting \(t=1\) gives \(\beta'\), and \(K(0,t)=K(1,t)=(0,0)\).
Keep the first loop fixed with \(H(u,t)=\alpha(u)\). The resulting homotopy of concatenations is
At \(s=\frac12\), the first piece is \((0,0)\), since \(2s=1\) and \(1-2s=0\). The second piece is \((0,0)\), since \(2s-1=0\). At \(s=\frac34\), the second piece uses \(2s-1=\frac12\) and \(2-2s=\frac12\); hence its value is \((0,(1-2t)/4)\). This is \((0,1/4)\) at \(t=0\) and \((0,-1/4)\) at \(t=1\). At \(s=0\) and \(s=1\) the values are both \((0,0)\), for every \(t\). These endpoint and seam checks confirm the required homotopy from \(\alpha*\beta\) to \(\alpha*\beta'\).
Worked Example: Keeping the Homotopy Inside the Interval
Let \(X=I=[0,1]\) with its subspace topology from the standard topology on \(\mathbb{R}\), and use \(0\) as the base point. Define loops \(\alpha(s)=4s(1-s)\), \(\alpha'(s)=s(1-s)\), \(\beta(s)=2s(1-s)\), and \(\beta'(s)=0\). Each has endpoint values \(0\). The formulas \(H(s,t)=(4-3t)s(1-s)\) and \(K(s,t)=(1-t)2s(1-s)\) give endpoint-preserving homotopies between the respective loops.
To check that these maps take values in \(I\), use \(0\le s(1-s)\le\frac14\). Since \(1\le4-3t\le4\), we have \(0\le H(s,t)\le(4-3t)/4\le1\). Since \(0\le1-t\le1\), we have \(0\le K(s,t)\le2/4=\frac12\). The concatenated homotopy is
At the seam both pieces are zero. At \(s=\frac14\) and \(t=\frac12\), the first piece is \((4-\frac32)(\frac12)(\frac12)=\frac52\cdot\frac14=\frac58\). At \(s=\frac34\) and \(t=\frac12\), the second piece is \(\frac12\cdot2\cdot\frac12\cdot\frac12=\frac14\). Both values lie in \(I\). At \(t=0\) the pieces are those of \(\alpha*\beta\); at \(t=1\) they are those of \(\alpha'*\beta'\). The formulas therefore exhibit the claimed homotopy while remaining in the stated space.
Why the Shared Endpoint Matters
The fixed-endpoint conditions are not technical decoration. They guarantee \(H(1,t)=K(0,t)\), the equality needed to paste the two halves. Without a common endpoint at every homotopy time, the proposed piecewise map may fail to be continuous at the seam.
Common pitfall. Suppose in \(\mathbb{R}\) one tries to concatenate \(p(s)=s\), which runs from \(0\) to \(1\), with \(q(s)=2+s\), which runs from \(2\) to \(3\). The usual piecewise formula gives value \(p(1)=1\) at \(s=\frac12\), but values from the second piece approach \(q(0)=2\) as \(s\) approaches \(\frac12\) from the right. Since \(1\ne2\), the two pieces do not meet continuously. The correction is to require the end of the first path to equal the start of the second; for based loops in this theorem, endpoint-preserving homotopies ensure that both seam values are always \(x_0\).
Thus the concatenation operation on based-loop homotopy classes is well-defined: choosing different endpoint-preserving homotopic representatives cannot change the resulting class. The proof depends only on constructing the piecewise homotopy and checking its continuity and endpoints.
Check Your Understanding
For each question, identify the endpoint or seam condition that is being used.
- Write the piecewise formula for a homotopy from \(\alpha*\beta\) to \(\alpha'*\beta'\), given endpoint-preserving homotopies \(H\) and \(K\).
- At \(s=\frac12\), what are the two values of that piecewise formula, and why are they equal?
- In the interval example, verify directly that \(H(s,t)\le1\) for all \(s,t\in I\).
- Why does fixing only the endpoints at \(t=0\) and \(t=1\), without fixing them at intermediate \(t\), fail to establish the theorem?
- For paths \(p\) and \(q\) that are not loops, what endpoint equality is needed to define their usual concatenation continuously?