1.10 Homotopy invariance

Below the video you will find accompanying notes and some pre-class questions.

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:XX is a continuous map which is homotopic to the identity map idX:XX, then the induced map F:π1(X,x)π1(X,F(x)) is an isomorphism.
(1.26) Let Ft:XX 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]=[Fγ]
and Fγ is freely homotopic to γ via the free homotopy Fsγ. This is a free (rather than based) homotopy because the basepoint of Fsγ is δ(s). Using our results on basepoint dependence, this implies that [Fγ]=[δγδ1].
and that the map [γ][δγδ1] is an isomorphism π1(X,x)π1(X,F(x)). Since F[γ]=[Fγ], this tells us that F is an isomorphism.

(6.38) If F:XY and G:YX 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 FG:YY is homotopic to idY. By the previous lemma, (FG):π1(Y,y)π1(Y,F(G(y))) is an isomorphism. By functoriality, we have (FG)=FG, so FG is an isomorphism. This implies that G is injective (otherwise FG would fail to be injective) and F is surjective (otherwise FG would fail to be surjective).

(9.58) By arguing the same way about the composition GF, 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

  1. The punctured plane C{0} is homotopy equivalent to the (1-dimensional) unit circle S1. Find ``1-dimensional'' topological spaces homotopy equivalent to Cμn 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(Cμn) once we have proved Van Kampen's theorem.

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