A description/proof of that for covering map, lift of product of paths is product of lifts of paths
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of covering map.
- The reader knows a definition of lift of continuous map with respect to covering map.
- The reader admits the proposition that for any covering map, there is the unique lift of any path for each point in the covering map preimage of the path image of any point on the path domain.
Target Context
- The reader will have a description and a proof of the proposition that for any covering map, the lift of the product of any 2 paths such that the product exists is the product of the lifts of the paths such that the product (of the lifts) exists.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any connected and locally path-connected topological spaces, \(T_1, T_2\), any covering map, \(\pi: T_1 \to T_2\), which means that \(\pi\) is continuous and surjective and around any point, \(p \in T_2\), there is a neighborhood, \(N_p \subseteq T_2\), that is evenly covered by \(\pi\), the closed interval, \(T_3 := I = [0, 1]\), any paths, \(f_1: T_3 \to T_2\) and \(f_2: T_3 \to T_2\) such that \(f_1 (1) = f_2 (0)\), and each point \(p'_0 \in \pi^{-1} (f_1 (0))\), the lift of the product, \(\widetilde{f_1 f_2}: T_3 \to T_1\) such that \(\widetilde{f_1 f_2} (0) = p'_0\), is the product of the lifts, \(\widetilde{f_1} \widetilde{f_2}\) such that \(\widetilde{f_1} (0) = p'_0\) and \(\widetilde{f_2} (0) = \widetilde{f_1} (1)\), which is \(\widetilde{f_1 f_2} = \widetilde{f_1} \widetilde{f_2}\).
2: Proof
There is the unique lift of \(f_1 f_2\), \(\widetilde{f_1 f_2}\), such that \(\widetilde{f_1 f_2} (0) = p'_0\), by the proposition that for any covering map, there is the unique lift of any path for each point in the covering map preimage of the path image of any point on the path domain. \(\pi \circ \widetilde{f_1 f_2} = f_1 f_2\). \(\pi \circ \widetilde{f_1 f_2} (2^{-1} r) = f_1 (r)\) for \(0 \le r \le 1\) and \(\pi \circ \widetilde{f_1 f_2} (2^{-1} + 2^{-1} r) = f_2 (r)\) for \(0 \le r \le 1\). In fact, \(\widetilde{f_1 f_2} (2^{-1} r)\) is the unique lift of \(f_1\) such that \(\widetilde{f_1} (0) = \widetilde{f_1 f_2} (0) = p'_0\), by the proposition that for any covering map, there is the unique lift of any path for each point in the covering map preimage of the path image of any point on the path domain and \(\widetilde{f_1 f_2} (2^{-1} + 2^{-1} i)\) is the unique lift of \(f_2\) such that \(\widetilde{f_2} (0) = \widetilde{f_1 f_2} (2^{-1}) = \widetilde{f_1} (1)\), by the proposition that for any covering map, there is the unique lift of any path for each point in the covering map preimage of the path image of any point on the path domain. The product, \(\widetilde{f_1} \widetilde{f_2}\), exists, and equals \(\widetilde{f_1 f_2}\).
3: Note
Lift is uniquely determined only after an initial value is specified. So, the product of the lift of \(f_1\) and an arbitrary lift of \(f_2\) may not exist; the product, \(\widetilde{f_1} \widetilde{f_2}\), exists because \(\widetilde{f_2}\) is chosen such that \(\widetilde{f_2} (0) = \widetilde{f_1} (1)\).
