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May 28, 2015 at 17:08 comment added Dylan Thurston They address it, but don't really go into the proof (as is sensible for that point in the book).
May 28, 2015 at 16:57 comment added Ian Agol I guess they address the issue of homotopy/homotopy equivalence vs. diffeomorphism/isotopy in Chapter 1.
May 27, 2015 at 21:43 comment added Ian Agol The mapping class group is usually defined as diffeos. modulo isotopies. To show equivalence with $Out(\pi_1(M))$, first one may show equivalence with self-homotopy equivalences up to homotopy, and then show that homotopy equivalences up to homotopy are equivalent to diffeomorphisms up to isotopy.
May 27, 2015 at 20:48 comment added Dylan Thurston Ian, can you be more explicit about exactly which statement you are arguing for? Those statements are both specific to surfaces, right?
May 27, 2015 at 20:45 comment added Ian Agol As for homotopy equivalence implies homeomorphism, one may prove this by induction on a hierarchy, a la Waldhausen. Similarly for homotopy equivalent homeos. are isotopic.
May 27, 2015 at 18:32 comment added Dylan Thurston Thanks, everyone. I wish authors would spell this out a bit more.
May 27, 2015 at 18:29 comment added mme $[X,K(G,n)] = H^n(X;G) = \text{Hom}(H_n(X;\Bbb Z);G) = \text{Hom}(\pi_n(X);G)$. For a more down-to-earth example of a map $X \to K(\pi,n)$ that's not null-homotopic but kills $\pi_n$ we could take the fibration $S^1 \to \Bbb{RP}^\infty \to \Bbb{CP}^\infty$ obtained by considering $\Bbb{RP}^\infty$ as a quotient of the infinite "odd-dimensional sphere" and extending this to a quotient by $S^1$. If this map was null-homotopic so would be the identity map of $\Bbb{RP}^\infty$. This should correspond to the Bockstein $H^1(-,\Bbb Z/2) \to H^2(-,\Bbb Z)$.
May 27, 2015 at 18:27 comment added mme I think you need some connectivity conditions on the domain - it seems to me there is a nontrivial map $K(\Bbb Z_2,1) \to K(\Bbb Z_2, 2)$ given by the cup square. (It is a theorem that there is a bijection between homotopy classes of maps $K(\pi, n) \to K(\pi', m)$ and natural transformations between the Set-valued functors $H^n(-, \pi) \to H^m(-,\pi')$. The cup square represents a nontrivial natural transformation because it does so on $\Bbb{RP}^2$, say.) If you assume the domain is $(n-1)$-connected you can prove what you want by exploiting the isomorphisms...
May 27, 2015 at 18:13 comment added Ryan Budney Correct. Things get complicated when your spaces have two or more non-trivial stages in their Postnikov towers.
May 27, 2015 at 18:00 comment added Dylan Thurston Thanks! I was looking in completely the wrong section of Hatcher. I guess the generally false belief is true more generally as long as the domain is a CW complex and the target is a $K(\pi, n)$, right?
May 27, 2015 at 17:51 comment added mme @DylanThurston: The generally false belief is true for based homotopy classes of maps $K(\pi, 1) \to K(\pi',1)$, which are in bijection with homomorphisms $\pi \to \pi'$. The proof Ryan alluded to is Hatcher's Theorem 1B.9. This theorem immediately implies that the homomorphisms induced by homotopy equivalences are precisely the automorphisms, and we pass to $\text{Out}(\pi,1)$ when we forget about the basepoint.
May 27, 2015 at 17:35 comment added Dylan Thurston Ryan, you're starting to sketch an argument for surjectivity of the map to endomorphisms of $\pi_1$. What about injectivity? Is, for instance, the false belief that I mentioned true for self-maps of $K(\pi,1)$ spaces, or for homotopy equivalences? This does fit within obstruction theory, and I looked a little in Hatcher's book before posting, but I couldn't find anything directly relevant (in terms of finding homotopies). I'd appreciate a more detailed pointer.
May 27, 2015 at 17:19 comment added Ryan Budney The link between continuous maps of a $K(\pi,1)$ and homomorphisms of $\pi_1$ up to conjugation comes from obstruction theory. Given a homomorphism of $\pi_1$ you define a map between the $K(\pi,1)$ and itself, inducing the map on $\pi_1$, cellularly. You can extend the map to the entire space since it has no higher homotopy groups. I think there is a semi-detailed write up in Hatcher's algebraic topology book. The same kind of argument shows you any continuous map between $K(\pi,1)$ spaces is induced via this kind of construction.
May 27, 2015 at 17:10 history asked Dylan Thurston CC BY-SA 3.0