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。 