命题(覆叠映射诱导的映射是单射的):
令 为一个覆叠空间,令 。那么诱导映射
是单射的。
证明: 假设 是在 处的环路,使得 。 那么存在一个从 到 的同伦,它们都是 处的环路。 这个同伦 可以唯一地提升 到一个同伦 ,使得 。 此外, 固定基点,因为映射 和 是连续的,因此将连通集映射到连通集,并且覆盖映射下一点的反像具有离散拓扑。 根据 路径提升的唯一性, 将等于 ,使得 。
证明:假设 提升 ,使得 。 那么 ,因此,由于 将 映射到 ,我们确实有 。
Conversely, suppose that . Then we define a lift of as follows: For each , choose a path so that and by path-connectedness of . By path lifting, lifts uniquely to a path . Then set . We have to show that this definition does not depend on the choice of . Indeed, let be another path like . Then is a loop at that induces an equivalence class . This in turn induces an equivalence class ; indeed, , since both are composition with the map . By the assumption, there exists a loop in such that . Moreover, is a path in that may be lifted to a path in . Since we may lift homotopies, we may lift a homotopy between and to a homotopy between and , which, similarly to the proof of injectivity of , leaves the endpoints fixed. Hence, is a loop, and in particular restricted to yields, when direction is reversed, a lift of that connects to . We conclude well-definedness.
It remains to prove continuity. Hence, let be arbitrary; we shall prove continuity at . Pick an evenly covered neighbourhood about . Let be the open set mapping homeomorphically to which contains . Let be any open neighbourhood of . Set to be the image of (which is open) under , so that is itself open. By continuity of , there exists an open neihbourhood of that is mapped by into . By strong local connectedness, choose to be connected. Recall that maps from connected domains lift uniquely; this shows that on , we have . Hence, maps into .
最后, 的连通性意味着 是唯一的。