1.10 Homotopy invariance
Below the video you will find accompanying notes and some pre-class questions.
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Notes
Homotopy invariance of the fundamental group
(0.00) In this section, we will prove that homotopy equivalent spaces have isomorphic fundamental groups.
(0.20) If F:X→X is a continuous map which is
homotopic to the identity map idX:X→X, then the
induced map F∗:π1(X,x)→π1(X,F(x)) is an
isomorphism.
(1.26) Let Ft:X→X be a homotopy from F0=F to
F1=idX and let δ be the path traced out by the
basepoint x under this homotopy, that is δ(t)=Ft(x).
Now, for [γ]∈π1(X,x), we have
F∗[γ]=[F∘γ]
and F∘γ is freely
homotopic to γ via the free homotopy
Fs∘γ. This is a free (rather than based) homotopy
because the basepoint of Fs∘γ is δ(s). Using
our results on basepoint dependence, this implies that
[F∘γ]=[δ⋅γ⋅δ−1].
and that the
map [γ]→[δ⋅γ⋅δ−1] is an
isomorphism π1(X,x)→π1(X,F(x)). Since
F∗[γ]=[F∘γ], this tells us that F∗ is an
isomorphism.
(6.38) If F:X→Y and G:Y→X are homotopy
inverses then F∗:π1(X,x)→π1(Y,F(x)) and
G∗:π1(Y,y)→π1(X,G(y)) are isomorphisms.
(7.36) The composition F∘G:Y→Y is homotopic to
idY. By the previous lemma, (F∘G)∗:π1(Y,y)→π1(Y,F(G(y))) is an isomorphism. By
functoriality, we have (F∘G)∗=F∗∘G∗, so F∗∘G∗ is an isomorphism. This implies that G∗ is injective
(otherwise F∗∘G∗ would fail to be injective) and F∗
is surjective (otherwise F∗∘G∗ would fail to be
surjective).
(9.58) By arguing the same way about the composition G∘F, we get that G∗ is surjective and F∗ is injective. This implies that both F∗ and G∗ are bijections, and hence isomorphisms.
Pre-class questions
- The punctured plane C∖{0} is homotopy
equivalent to the (1-dimensional) unit circle S1. Find
``1-dimensional'' topological spaces homotopy equivalent to
C∖μn where μn is the set of nth
roots of unity (a doodle, rather than a proof, will suffice). We
will be able to use this to compute
π1(C∖μn) once we have proved Van Kampen's
theorem.
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