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Let $\ M'\ M''\ $ be simply-connected Hausdorff compact manifolds (possibly with boundary for another variant of the question). Let $\ f:M'\rightarrow M''\ $ be a continuous function which induces an isomorphism of the cohomological rings. Does there exist a continuous function $\ g:M''\rightarrow M'\ $ which induces the inverse cohomology ring isomorphism?

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    $\begingroup$ Should be true (at least if the manifolds are smooth). Morse theory shows that the manifolds have CW-structures. Then Whitehead's theorem states that $f$ is a homotopy equivalence, hence there is an inverse homotopy equivalence. As you probably noticed, homology spheres show that simply-connected is necessary. $\endgroup$ – Matthias Wendt Aug 21 '14 at 6:03
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    $\begingroup$ Smoothness is not necessary; every (paracompact Hausdorff) topological manifold has the homotopy type of a CW-complex. $\endgroup$ – Eric Wofsey Aug 21 '14 at 6:11
  • $\begingroup$ Thank you Matthias and Eric. Furthermore, instead of simple-connectedness I could assume that $\ f\ $ induces also an isomorphism of the fundamental groups. $\endgroup$ – Włodzimierz Holsztyński Aug 21 '14 at 6:16
  • $\begingroup$ @Matthias: I was trying to see if homology spheres gave a counterexample without the hypothesis of simple connectedness but I couldn't finish it. Can you give some details? $\endgroup$ – Qiaochu Yuan Aug 21 '14 at 16:27
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    $\begingroup$ @QiaochuYuan: here is the argument I had in mind. Take an integral homology 3-sphere $M$ with contractible universal cover. Such things can be found among the Brieskorn manifolds. Then the map to the plus-construction $M\to M^+\simeq S^3$ is a continuous map which induces an isomorphism in cohomology. But any map $S^3\to M$ must have degree 0, so there can not be an inverse on cohomology. $\endgroup$ – Matthias Wendt Aug 21 '14 at 17:44
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The question has been answered in the comments. To justify writing this answer I'll sweeten it with some links and further details.

Compact manifolds (possibly with boundary) have the homotopy type of a CW-complex, see the answers to this MO-question which also provide links to the literature.

If $f:X\to Y$ induces isomorphisms on cohomology and $X$ and $Y$ are simply-connected spaces, then $f$ is a weak equivalence, discussed e.g. here. Subtleties with homological Whitehead theorems in the non-simply-connected case are discussed in this MSE-question. I suppose that the implication from cohomology isomorphism to weak equivalence needs isomorphisms on fundamental groups and cohomology with all local systems coefficients, unless the space is nilpotent, see this paper of Gersten and the cited paper of E. Dror.

If $f:X\to Y$ is a weak equivalence of CW-complexes, then it is a homotopy equivalence, and therefore has a homotopy inverse $g:Y\to X$, which induces the inverse cohomological isomorphism for $f^\ast:H^\ast(Y)\to H^\ast(X)$. By the first remark this applies to all compact manifolds with boundary.

Finally, an example with homology spheres, some other examples of what can go wrong have been provided in the links above. In Milnor's paper on Brieskorn homology spheres, one can find lots of examples of homology spheres which have a contractible universal cover (they are covered by hyperbolic three-space and their fundamental groups are related to triangle groups). The plus-construction again provides a map $M\to S^3$ from the homology sphere to $S^3$. (The plus-construction produces a CW-complex which by Whitehead's theorem again is homotopy equivalent to $S^3$.) This map $M\to S^3$ induces an isomorphism on cohomology with integral coefficients. However, there can not be an inverse, because the universal covering of $M$ is contractible and so every map $S^3\to M$ has degree $0$. This provides another example that some condition like simply-connected is necessary.

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  • $\begingroup$ Correction: The universal covering map $S^3 \to SO(3)/I$ has degree 120, the order of the fundamental group of $SO(3)/I$. $\endgroup$ – Allen Hatcher Sep 4 '14 at 14:07
  • $\begingroup$ @AllenHatcher: Oh, yes, sorry. I deleted that. Is there a degree $1$ map $S^3\to SO(3)/I$, or is it possible to show that there is no such map? $\endgroup$ – Matthias Wendt Sep 4 '14 at 14:14
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Some further comments on homology spheres: First, for any closed connected orientable $n$-manifold $M$ there is always a degree $1$ map $M\to S^n$ obtained by collapsing the complement of an open ball in $M$ to a point. In the reverse direction, any degree $1$ map $f:M\to N$ of closed connected orientable $n$-manifolds must induce a surjection on $\pi_1$, for otherwise $f$ could be lifted to the covering space $\tilde N \to N$ corresponding to the proper subgroup $f_*(\pi_1M) \subset \pi_1N$. If this covering is finite-sheeted, then the degree of the projection $\tilde N\to N$ is equal to the number of sheets (which is the index of $f_*(\pi_1M)$ in $\pi_1N$) since an orientation of $N$ lifts to an orientation of $\tilde N$, making the local degrees at all points in a fiber have the same sign. Thus the degree of $f$ is divisible by the number of sheets, so it can't be $1$. If the covering is infinite-sheeted then $\tilde N$ is noncompact so $H_n(\tilde N)=0$, forcing $f$ to have degree $0$. Applying this fact to a degree $1$ map $S^n\to N$ we see that $\pi_1N=0$ so it can't be a nonsimply-connected homology sphere. (These are classical arguments, incidentally.)

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    $\begingroup$ You mean $H_n(\tilde{N}) = 0$? $\endgroup$ – Lennart Meier Sep 4 '14 at 23:41
  • $\begingroup$ @Lennart Meier: Yes, thanks. Corrected now. $\endgroup$ – Allen Hatcher Sep 5 '14 at 4:25

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