Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Today my fellow grad student asked me a question, given a map f from X to Y, assume $f_*(\pi_i(X))=0$ in Y, when is f null-homotopic?

I search the literature a little bit, D.W.Kahn

http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102995805

And M.Sternstein has worked on this, and Sternstein even got a necessary and sufficient condition, for suitable spaces.

http://www.jstor.org/stable/pdfplus/2037939.pdf

However, his condition is a little complicated for me as a beginner. Right now I just wanted a counter example of a such a map. Kahn in his paper said one can have many such examples using Eilenberg Maclance spaces. Well, we can certainly show a lot of map between E-M spaces induce zero map on homopoty groups just by pure group theoretic reasons, but I can not think of a easy example when you can show that map, if it exists, is not null-homotopic. Could someone give me some hint?

or, maybe even some examples arising from manifolds?

share|improve this question
add comment

4 Answers

up vote 15 down vote accepted

Consider ordinary singular cohomology with varying coefficients. You can look at the short exact sequence of abelian groups:

$$0 \to \mathbb{Z}/2 \to \mathbb{Z}/4 \to \mathbb{Z}/2 \to 0$$

This gives rise, for any space X, to a short exact sequence of chain complexes:

$$0 \to C^i(X;\mathbb{Z}/2) \to C^i(X;\mathbb{Z}/4) \to C^i(X;\mathbb{Z}/2) \to 0$$

and hence you get a long exact sequence in cohomology. Thus we get an interesting boundary map known as the Bockstein

$$H^i(X; \mathbb{Z}/2) \to H^{i+1}(X; \mathbb{Z}/2).$$

This is natural in X and so is represented by a (homotopy class of) map(s) of Eilenberg-Maclane spaces:

$$K(i, \mathbb{Z}/2) \to K(i+1, \mathbb{Z}/2)$$

This map is necessarily zero on homotopy groups. To show that this map is not null-homotopy, you just need to find a space for which the Bockstein is non-trivial. There are lots of examples of this. Rather then explain one, I suggest you look up "Bockstein homomorphism" in a standard algebraic topology reference, e.g. Hatcher's book.

share|improve this answer
    
Thanks, this argument is very nice! I guess it is not very hard to find a space with non trivial Bockstein, e.g.$M=RP^2$, we have $H^i(M,Z/2)=H^i(M,Z/4)=Z/2$for i=1,2, and if I look at the reduced cohomology long exact sequence $H^0(M,pt,Z/2)=H^0(M,pt,z/4)=0$. Then the Bockstein from $H^2$to $H^1$will not be trivial, otherwise we will have $0\to Z/2\to Z/2\to Z/2\to 0$ contradiction. Although I used the reduced cohomology, since I only use maps to K(z/2,2) and K(Z/2,1) to represent $H^1$ and $H^2$ I am fine. From this example, we get a non null-homotopic map from $CP^{\infty}$ to $RP^\infty$! –  Ying Zhang Apr 4 '10 at 4:05
    
Sorry about my last sentence, from the above sequence I got a map from $RP^{\infty}\to K(Z/2,2)$. To get a map from $RP^{\infty}\to CP^{\infty}$, we should look at the coefficient sequence $0\to Z\to Z\to Z/2Z\to 0$ which does the job. Geometrically it is not too hard to find an interesting map from $RP^{\infty}$ to $CP^{\infty}$ just by quotient out more things. –  Ying Zhang Apr 4 '10 at 18:15
add comment

For a more explicit example than Chris's, consider the map from the (2-dimensional) torus to a sphere that collapses the 1-skeleton of the usual CW complex and takes the 2-cell to the 2-cell of the sphere. The torus is $K(1, \mathbb{Z}^2)$, so this necessarily gives zero maps on homotopy, but it's also pretty clearly not null-homotopic.

share|improve this answer
    
Nice! So this will give us examples when X and Y are both manifolds:) –  Ying Zhang Apr 4 '10 at 4:11
add comment

Even if you ask that $f$ induces trivial maps on all (singular) homology and cohomology groups, there are still easy manifold examples. (This actually arises as an exercise in Hatcher's AT).

For instance, let $f:T^3\rightarrow S^2$ be the composition $T^3\rightarrow S^3\rightarrow S^2$, where the map from $T^3$ to $S^3$ is simply collapsing the 2-skeleton to a point, and the map from $S^3$ to $S^2$ is the Hopf map.

As others have mentioned, since $T^3$ is a $K(\mathbb{Z}^3, 1)$, if follows that $f$ induces trivial maps on homotopy groups.

Since the Hopf map induces trivial maps on homology and cohomology, it follows that $f$ does as well.

Finally, to see that $f$ is NOT nullhomotopic, assume it is. Since the map from $S^3$ to $S^2$ is a fiber bundle, it has the homotopy lifting property. Hence, we can lift the homotopy of $f$ to a homotopy $G:I\times T^3\rightarrow S^3$ where $G_0$ is the above map from $T^3$ to $S^3$ and $G_1$ is is a map from $T^3$ to $S^1\subseteq S^3$, the preimage of a point in $S^2$ under the Hopf map.

But $G_0$ has degree 1, while $G_1$ has degree 0, a contradiction.

share|improve this answer
    
Thanks. Btw, the exercise you mentioned is on Hatcher p.392 problem 34. –  Ying Zhang Apr 4 '10 at 18:18
add comment

I asked a very similar question a few months ago, and got some excellent answers.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.