Show that composition of paths satisfies the following cancellation property: If \(f_0 \cdot g_0 \simeq f_1 \cdot g_1\) and \(g_0 \simeq g_1\) then \(f_0 \simeq f_1\).
Solution
1. Properties of the path homotopy groupoid:
1.1. Let \([f]\) denote the path homotopy class of a path \(f: [0, 1] \to X\) relative to its endpoints \(\{0, 1\}\).The path concatenation \([f] \cdot [g] = [f \cdot g]\) is well-defined when \(f(1) = g(0)\).
standard definition of path concatenation and homotopy classes.
1.2. Path concatenation satisfies:- Associativity: \(([f] \cdot [g]) \cdot [h] = [f] \cdot ([g] \cdot [h])\) whenever the compositions are defined.- Right Identity: \([f] \cdot [c_{f(1)}] = [f]\), where \(c_x\) denotes the constant path at \(x\).- Right Inverse: For the reverse path \(\bar{g}(s) = g(1 - s)\), \([g] \cdot [\bar{g}] = [c_{g(0)}]\).
fundamental groupoid axioms.
2. Algebraic derivation of the cancellation:
2.1. By hypothesis:\[ [f_0 \cdot g_0] = [f_1 \cdot g_1] \implies [f_0] \cdot [g_0] = [f_1] \cdot [g_1]. \]
hypothesis \(f_0 \cdot g_0 \simeq f_1 \cdot g_1\).
2.2. Since \(g_0 \simeq g_1\), we have \([g_0] = [g_1]\), which also gives \([\bar{g}_0] = [\bar{g}_1]\).
reversing homotopies between paths preserves path homotopy classes.
2.3. Multiply both sides of 1 on the right by \([\bar{g}_0]\):\[ ([f_0] \cdot [g_0]) \cdot [\bar{g}_0] = ([f_1] \cdot [g_1]) \cdot [\bar{g}_0]. \]
well-definedness of multiplication in the path groupoid.
2.4. Applying associativity and substituting \([\bar{g}_0] = [\bar{g}_1]\) on the right-hand side:\[ [f_0] \cdot ([g_0] \cdot [\bar{g}_0]) = [f_1] \cdot ([g_1] \cdot [\bar{g}_1]). \]
associativity and 2.
2.5. Simplifying the inverse pairs:Since \([g_0] \cdot [\bar{g}_0] = [c_{g_0(0)}] = [c_{f_0(1)}]\) and \([g_1] \cdot [\bar{g}_1] = [c_{g_1(0)}] = [c_{f_1(1)}]\):\[ [f_0] \cdot [c_{f_0(1)}] = [f_1] \cdot [c_{f_1(1)}] \implies [f_0] = [f_1]. \]Thus \(f_0 \simeq f_1\).
right identity property of constant paths.
3. Conclusion:\(f_0 \simeq f_1\). Q.E.D.
2.