Let $X$ and $Y$ be connected topological spaces. Can we then realize each map between their homologies as coming from a continuous map between their spaces?

For any arrow $φ\colon \tilde H_•(X) → \tilde H_•(Y)$ in $\mathrm{Ab}^ℤ$, is there an arrow $f\colon X → Y$ in $\mathrm{Top}$ such that $\tilde H_•(f) = φ$? How about if $X = Y$? Are there any nice conditions under which this is true?

  • $\begingroup$ math.stackexchange.com/questions/1091709/… $\endgroup$ – Moishe Kohan Jan 29 '15 at 19:35
  • $\begingroup$ @studiosus Thanks. I guess you could make that an answer here as well. Do you also have a counterexample for when $X = Y$? $\endgroup$ – k.stm Jan 29 '15 at 19:37
  • $\begingroup$ +1 for asking the same question in the title and the body! (Some people have a habit of asking "Q?" in the title and "not-Q?"in the body.) $\endgroup$ – Qiaochu Yuan Jan 29 '15 at 20:00
  • $\begingroup$ @QiaochuYuan Thanks for giving me credit for this – but out of curiosity: How is that habit bad? $\endgroup$ – k.stm Jan 29 '15 at 21:21
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    $\begingroup$ It means that "yes" and "no" become ambiguous. It's not a big deal, it's just slightly annoying. $\endgroup$ – Qiaochu Yuan Jan 29 '15 at 22:25

No. There is an obstruction coming from the cup product, and so it will be easier to talk about cohomology. A simple family of examples when $X = Y$ is to take $X = Y = \mathbb{CP}^n$ for $n \ge 2$. As a graded ring with the cup product,

$$H^{\bullet}(\mathbb{CP}^n) \cong \mathbb{Z}[x]/x^{n+1}$$

where $x$ is the generator of $H^2(\mathbb{CP}^n)$, living in degree $2$. It follows that the action of a map $f : X \to Y$ on cohomology (and hence on homology) is completely determined by the map $H^2(f) : H^2(Y) \to H^2(X)$, which sends $x$ to $mx$ for some integer $m$. Compatibility with the cup product forces $H^{2k}(f) : H^{2k}(Y) \to H^{2k}(X)$ to send $x^k$ to $m^k x^k$. Dualizing gives the corresponding statement for the induced map on homology as well.

There are other subtler obstructions coming from more exotic cohomology operations, but over $\mathbb{Z}$ these are a bit hard to work with. In particular, even if the cup product vanishes there are still obstructions on the induced map on $\mathbb{Z}_2$ cohomology (which indirectly obstructs the induced map on $\mathbb{Z}$ homology) coming from the fact that it has to respect Steenrod operations.

  • $\begingroup$ (The question was about homology :-) ) $\endgroup$ – Mariano Suárez-Álvarez Jan 30 '15 at 22:53
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    $\begingroup$ @MarianoSuárez-Alvarez “Dualizing gives the corresponding statement for the induced map on homology as well”. $\endgroup$ – k.stm Jan 31 '15 at 7:19
  • $\begingroup$ I think Mariano was joking. (Maybe?) $\endgroup$ – Qiaochu Yuan Jan 31 '15 at 10:25

It is nice to see so many elaborate examples, but here is a trivial one...

Let $X$ be a space with one point. There is no continuous function $f:X\to X$ such that the map $f:H_\bullet(X)\to H_\bullet(X)$ is multiplication by $2$.

Later. The above example is for unreduced homology homology, but a similar sort of example (in that the obstruction comes simply from counting) can be used easily. Consider the poset

enter image description here

and endow the four point set with the order topology, and call it $X$. The space $X$ has the weak homotopy type of $S^1$, and therefore $H_1(X)\cong\mathbb Z$. There is a finite number of maps $X\to X$, simply because $X$ has only four points, and there are infinitely many maps $H_\bullet(X)\to H_\bullet(X)$, so some of the latter cannot be gotten from maps of the space.

Indeed, given any compact CW-complex there is a finite poset with the same weak homotopy type, and we can do the same silly counting argument to get examples.

  • $\begingroup$ Uh? $\tilde H(pt)=0$. $\endgroup$ – Grigory M Jan 30 '15 at 23:19
  • $\begingroup$ Take two points for reduced homology. $\endgroup$ – Mariano Suárez-Álvarez Jan 30 '15 at 23:20
  • $\begingroup$ The question is about the category of connected spaces. $\endgroup$ – Grigory M Jan 30 '15 at 23:21
  • $\begingroup$ Ah, right. ${}$ $\endgroup$ – Mariano Suárez-Álvarez Jan 30 '15 at 23:21
  • $\begingroup$ There you go. :-) $\endgroup$ – Mariano Suárez-Álvarez Jan 30 '15 at 23:37

Here are some aspherical examples.

Suppose that $M$ is a closed hyperbolic (real or complex) manifold with 1-st Betti number $>1$ (such exist in abundance); I will assume that the real dimension of $M$ is $>2$. Then the group of outer automorphisms of $\pi_1(M)$ is finite (by the Mostow Rigidity Theorem), while the group of automorphisms of the 1st homology group is infinite. The outer automorphism group $Out(\pi_1(M))$ is the same as the group of self-homotopy-equivalences of $M$ due to asphericity of $M$.

To make things more interesting, take a 4-dimensional closed real-hyperbolic manifold with $b_2(M)>0$ (examples, again, abound). Since $M$ has zero signature, the group of automorphisms of the intersection form on $H_2(M)$ is infinite, while $Out(\pi_1(M))$ is again finite.

Edit. Here are some details:

Definition. A group $G$ is called Hopfian if each epimorphisms $G\to G$ is an isomorphism.

Malcev proved in 1940 that all finitely-generated matrix groups are Hopfian. (He noticed that each residually finite group is Hopfian and then proved that all finitely generated matrix groups are residually finite.) I will use this in the context of fundamental groups of complete hyperbolic and complex-hyperbolic manifolds. These groups are subgroups of $PO(n,1)$ and $PU(n,1)$ respectively (where $n$ is the real and complex dimension respectively). For instance, if $M$ is a hyperbolic n-manifold, then $M=H^n/G, G<PO(n,1)$, $H^n$ is the hyperbolic n-space.

Lemma. Suppose that $M$ is a closed oriented manifold with Hopfian fundamental group. Let $f: M\to M$ be a degree $\pm 1$ map. Then $f$ induces an automorphism of $\pi_1(M)$.

Proof. Let $G=\pi_1(M)$, $H=f_*(G)<G$. If $H\ne G$, then the map $f$ lifts to a map $f': M\to M'$, where $p: M'\to M$ is the covering corresponding to the subgroup $H$. Since $deg(f)=deg(f')deg(p)$ (with both sides equal to zero if $p$ has infinite degree), we obtain a contradiction. Hence, $H=G$. Since $G$ is Hopfian, $f_*$ is an isomorphism. qed

Corollary. If $M$ is a real or complex-hyperbolic manifold, then any degree 1 map $f: M\to M$ is a homotopy-equivalence.

Proof. Since $M$ is aspherical (its universal cover is the unit ball), and $f_*: \pi_1(M)\to \pi_1(M)$ is an isomorphism, $f$ is a homotopy-equivalence by Whitehead's theorem. qed

Remark. It is an open problem, first posed by Hopf himself, if any degree 1 map $M\to M$ of a closed oriented topological manifold is a homotopy equivalence, see here.

Corollary. In the above setting, assuming that real dimension of $M$ is at least 3, there are only finitely many automorphisms $\phi: H_*(M)\to H_*(M)$ induced by continuous self-maps $f: M\to M$.

Proof. Suppose that $\phi$ is such an automorphism. Then, by looking at the action on the top-dimensional homology, $f$ has to be a homotopy-equivalence. Mostow proved (Mostow rigidity theorem) that each homotopy-equivalence $f: M\to M$ is homotopic to an isometry. (Wikipedia article states this only for hyperbolic manifolds, but Mostow proved much more!) Lastly, the isometry group of a closed real or complex-hyperbolic manifold (of any dimension $>1$) is finite. This follows, for instance, from the Arzela-Ascoli's theorem and the fact that an isometry homotopic to the identity has to be the identity map. qed

  • $\begingroup$ I don't think this answers the question. You're showing that homology isn't surjective on isomorphisms, which does not imply that it isn't full. $\endgroup$ – Qiaochu Yuan Jan 29 '15 at 20:11
  • $\begingroup$ For a simpler example, there are maps of rings which are surjective but which aren't surjective on groups of units, such as $\mathbb{Z} \to \mathbb{Z}_5$. $\endgroup$ – Qiaochu Yuan Jan 29 '15 at 20:17
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    $\begingroup$ @QiaochuYuan: Yes, it does answer the question. The point is that any degree one self-map $f: M\to M$ ($M$ being real or complex hyperbolic) has to be a homotopy-equivalence. (This is not completely obvious, but not hard to prove once you know that $\pi_1$ is Hopfian.) $\endgroup$ – Moishe Kohan Jan 29 '15 at 21:25
  • $\begingroup$ Oh, interesting. Sorry for the downvote, then; I can't remove it anymore. $\endgroup$ – Qiaochu Yuan Jan 29 '15 at 23:46
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    $\begingroup$ @QiaochuYuan If studiosus edits his (her?) answer pro forma, you can remove the downvote. By the way, studiosus: This answer is on a level a tad too high for me – I don’t understand it yet, that’s why I cannot upvote it. But thanks anyway. $\endgroup$ – k.stm Jan 30 '15 at 6:58

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