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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

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

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?

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5 Answers 5

up vote 22 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.

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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

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 $K(\mathbb{Z}^2,1)$, so this necessarily gives zero maps on homotopy, but it's also pretty clearly not null-homotopic.

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Nice! So this will give us examples when X and Y are both manifolds:) – Ying Zhang Apr 4 '10 at 4:11
You can also generalize this to the case where X is an $n$-manifold and a $K(G,1)$ (or even just having nonzero homotopy groups only up to $n-1$), and $X \to S^n$ is the map that collapses everything outside a Euclidean ball to a point. – Kevin Casto May 27 at 23:26

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.

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Thanks. Btw, the exercise you mentioned is on Hatcher p.392 problem 34. – Ying Zhang Apr 4 '10 at 18:18

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

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I realize this is a very old question. Nonetheless, here is a large, easy, and well-known class of examples.

Let $X$ be a path connected CW-complex, and consider the diagonal map $\delta_X\colon X\to X\wedge X$ into the smash product (i.e., the composite of $X\xrightarrow{\text{diag}} X\times X \xrightarrow{\text{quot}} X\wedge X$).

This map is often non-null: for instance, $$ \widetilde{H}^*(X)\otimes \widetilde{H}^*(X)\xrightarrow{\text{Kunneth}} \widetilde{H}^*(X\wedge X)\xrightarrow{\delta_X} \widetilde{H}^*(X) $$ is exactly the cup-product, so any $X$ with non-trivial cup-product in positive degrees must have non-null $\delta_X$.

On the other hand, $\delta_X$ is always trivial on homotopy groups: any $f\colon S^k\to X$ fits in the commutative square $$\begin{array}{ccc} S^k & \xrightarrow{\delta_{S^k}} & S^k\wedge S^k \\ \downarrow & & \downarrow \\ X & \xrightarrow{\delta_X} & X\wedge X \end{array} $$ and $\delta_{S^k}\sim *$ for $k\geq 1$. In fact, this idea proves that any composite $\Sigma Y\to X\xrightarrow{\delta_X} X\wedge X$ is null, or equivalently that $\Omega(\delta_X)\colon \Omega X\to \Omega(X\wedge X)$ is null-homotopic for any $X$.

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Just because you didn't mention it explicitly: this is also one way to prove the cup product vanishes identically on spaces which are suspensions of other spaces. – Omar Antolín-Camarena Oct 9 at 17:50

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