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30 Chapter 1 The Fundamental Group
To show that Φ is injective, suppose Φ(m) = Φ(n), which means ωm ωn .
Let ft be a homotopy from ωm = f0 to ωn = f1 . By (b) this homotopy lifts to a
homotopy ft of paths starting at 0. The uniqueness part of (a) implies that f0 = ω m
and f1 = ω n . Since ft is a homotopy of paths, the endpoint ft (1) is independent
of t . For t = 0 this endpoint is m and for t = 1itisn,som=n.
It remains to prove (a) and (b).
general assertion:
Both statements can be deduced from a more
(c) Given a map F : Y ×I→S 1 and a map F : Y ×{0}→Rlifting F|Y ×{0}, then there
is a unique map F : Y ×I→R lifting F and restricting to the given F on Y ×{0}.
Statement (a) is the special case that Y is a point, and (b) is obtained by applying (c)
with Y = I in the following way. The homotopy f t in (b) gives a map F : I×I→S 1
by setting F(s,t) = ft (s) as usual. A unique lift F : I×{0}→Ris obtained by an
application of (a). Then (c) gives a unique lift F : I×I→R. The restrictions F|{0}×I
and F|{1}×I are paths lifting constant paths, hence they must also be constant by
the uniqueness part of (a). So ft (s) = F(s,t) is a homotopy of paths, and ft lifts ft since p F = F .
We shall prove (c) using just one special property of the projection p : R→S 1 ,
namely:
(∗)
There is an open cover {Uα } of S 1 such that for each α, p −1 (Uα ) can be
decomposed as a disjoint union of open sets each of which is mapped homeomorphically
onto Uα by p .
For example, we could take the cover {U α } to consist of any two open arcs in S 1
whose union is S 1 .
To prove (c) we will first construct a lift F : N ×I→R for N some neighborhood
in Y of a given point y0 ∈ Y . Since F is continuous, every point (y0 ,t) ∈ Y×I has
a product neighborhood Nt ×(at ,bt ) such that F Nt ×(at ,bt ) ⊂ Uα for some α.
By compactness of {y0 }×I, finitely many such products Nt ×(at ,bt ) cover {y0 }×I.
This implies that we can choose a single neighborhood N of y0 and a partition
0 = t0