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

Does anyone have good examples of a space $X$ and a map $f: X \to X$ so that $f_*: H_*(X) \to H_*(X)$ is the identity but (e.g.) $f_*: H_*(X; \mathbb{F}_2) \to H_*(X; \mathbb{F}_2)$ is not the identity?

Edit: As mentioned in the comments, $f_*$ is an isomorphism on the mod-2 homology, but I don't see why it needs to be the identity. More precisely, the exact sequence \[ C(X; \mathbb{Z}) \overset{\times 2}{\longrightarrow} C(X; \mathbb{Z}) \longrightarrow C(X;\mathbb{Z}/2) \] gives an exact triangle in homology, which in turn induces a 2-step filtration on $H_*(X; \mathbb{Z}/2)$ (where one step is the image of the map $H_*(X;\mathbb{Z}) \to H_*(X;\mathbb{Z}/2)$). The assumption that $f_*$ is the identity on integral homology implies that it is the identity on the associated graded space to this filtration, but that still doesn't imply it is the identity.

I came across a similar phenomenon in the context of Heegaard Floer homology, with the rings $(\mathbb{Z}/2)[U]$ and $\mathbb{Z}/2$ playing the roles of $\mathbb{Z}$ and $\mathbb{Z}/2$.

share|improve this question
Doesn't the UCT imply that no such counterexample exists? The splitting isn't natural, but the short exact sequence is. More simply, the multiplication by $2$ map on $\mathbb{Z}$ induces a LES in the homology of $X$ and if $f$ is a homology iso, the 5 lemma shows it is an iso with mod $2$ coefficients. Or am I missing something? –  Sam Isaacson Mar 29 '11 at 4:43
BTW, the simplest example of the failure of the UCT to split naturally that I know of is the map from a mod $n$ Moore space to a sphere that collapses the bottom cell. –  Sam Isaacson Mar 29 '11 at 4:46
@Sam: the question is: is that isomorphism also the identity? Notice that an endomorphism of short exact sequences which has identites on the left and on the right does not necessarily have an identity in the middle. –  Mariano Suárez-Alvarez Mar 29 '11 at 5:09
@Mariano, thanks for pointing out the subtlety I missed. I think you can produce a counterexample stably by looking at $X = M \wedge \Sigma M$ where $M$ is the mod $2$ Moore spectrum. Let $f:X \to X$ be the map which on $\Sigma M$ is the inclusion of the $\Sigma M$ summand and on $M$ is the sum of the inclusion of the $M$ summand and the essential composition $g:M\to S^1 \to \Sigma M$. The map $f$ is the identity on $H\mathbb{Z}$, but since $g$ is nonzero on $H\mathbb{Z}/2$, the map $f$ is not the identity on $H\mathbb{Z}/2$. –  Sam Isaacson Mar 29 '11 at 5:52
@Sam: If you take a mod-2 Moore space and suspend it, you should be able to do that unstably. –  Tyler Lawson Mar 29 '11 at 13:42
show 2 more comments

3 Answers

up vote 10 down vote accepted

To expand on my comment, let $M = \mathbf{R}P^2$ be the Moore space with (reduced) homology concentrated in dimension $1$. Let $f:M \to \Sigma M$ be the map $$ M \to S^2 \to \Sigma M $$ given by collapsing the $1$-skeleton of $M$ and then including the bottom cell into $\Sigma M$. This map induces $0$ on $\tilde H_\ast({-};\mathbf{Z})$ for dimension reasons. However, the map $f$ is an isomorphism on $H_2({-};\mathbf{Z}/2)$; this follows from the long exact sequences $$ \dotsb \to H_2(S^1;\mathbf{Z}/2) \to H_2(M;\mathbf{Z}/2) \to H_2(S^2;\mathbf{Z}/2) \to \dotsb $$ and $$ \dotsb \to H_2(S^2;\mathbf{Z}/2) \to H_2(\Sigma M;\mathbf{Z}/2) \to H_2(S^3;\mathbf{Z}/2) \to \dotsb .$$ This is an example of "non-naturality."

To obtain a self-map of a space $X$ that is the identity on $H_\ast(X;\mathbf{Z})$ but not on $H_\ast(X;\mathbf{Z}/2)$, we follow Tyler's suggestion: let $X = \Sigma M \vee \Sigma^2 M$. Since $X$ is a co-$H$-space, we can add maps in $[X,X]$. Let $g:X\to X$ be the sum of $1_X$ and the map $$ X \to \Sigma M \xrightarrow{\Sigma f} \Sigma^2 M \to X $$ where the first map collapses $\Sigma^2 M$ and the third map is the inclusion of $\Sigma^2 M$. The induced map in homology is $1 + \Sigma f_\ast$. In $H_\ast(X;\mathbf{Z})$, this is $1$. However, in $H_3(X;\mathbf{Z}/2)$, the map $g$ is not the identity, since $\Sigma f_\ast$ is nonzero. If we fix the basis of $H_3(X;\mathbf{Z}/2)$ given by the wedge decomposition of $X$, then $g_\ast$ is represented by the matrix

$$ \begin{pmatrix} 1 & 1 \\\\ 0 & 1 \end{pmatrix}. $$

share|improve this answer
Sam's two examples can be combined to make a slightly simpler answer to the question by letting $X$ be $S^2\vee M$. –  Tom Goodwillie Mar 29 '11 at 17:22
add comment

Concerning the analogous question for cohomology: There is a homeomorphism from the Klein bottle to itself that induces the identity map in integral cohomology but not in mod $2$ cohomology.

This example is Spanier-Whitehead dual to the $S^2\vee \mathbb RP^2$ variant of Sam's example.

For an example of a map inducing the identity in both integral homology and in integral cohomology, but not in mod $2$ (co)homology, you need to use a homology group that is not finitely generated.

In Sam's example (map from $\mathbb RP^2\vee \Sigma \mathbb RP^2$ to itself) the map is zero on both integral homology and integral cohomology.

share|improve this answer
add comment

Here's another solution; it's really the same as the ones Sam and Tom give, but with a more "geometric" flavor.

Start with the unit 2-sphere $S^2\subset \def\R\mathbb{R}\R^3$, and let $r: S^2\to S^2$ be given by reflection across the $xz$-plane ($r(x,y,z)=r(x,-y,z)$). This passes to a self-map $f: \R P^2\to \R P^2$ on the quotient; it carries the subspace $\R P^1\subset \R P^2$ to itself, which I'm thinking of as the quotient of the circle in the $xy$-plane. In other words, $f$ is a cellular map, with respect to the "usual" CW-filtration $\R P^0\subset \R P^1\subset \R P^2$. On the "cellular chain complex" of $\R P^2$, the map $f$ is degree $-1$ on the cells in dimension $1$ and $2$.

Now let $X=\R P^2\cup_{\R P^1} \R P^2$, obtained by gluing two projective planes along a circle. Let $g:X\to X$ be the self-map which (i) sends the first $\R P^2$ to the second $\R P^2$ by the map $f$, (ii) sends the second $\R P^2$ to the first $\R P^2$ by the map $f$, and thus (iii) sends the common $\R P^1$ to itself by the map $f|\R P^1$.

Since $g$ is a cellular map, its easy to compute its effect on the cellular chain complex of $X$: it's degree $-1$ on the $1$-cell, and it switches the two $2$-cells with each other by degree $-1$ maps. So it's identity on $H_*(X;\def\Z\mathbb{Z}\Z)$, since $H_1(X;\Z)=\Z/2$ and the "cellular chain" $(1,-1)$ which generates $H_2(X;\Z)=\Z$ is clearly fixed; but it's not the identity on $H_2(X;\Z/2)=\Z/2\oplus \Z/2$, as the switching of the $2$-cells is visible here.

(The generator of $H_2(X;\Z)$ is actually the image of a map $q:S^2\to X$, which you get by gluing the characteristic maps of the two $2$-cells together, and $fq$ differs from $q$ exactly by a 180-degree rotation of the sphere (and so $q$ and $fq$ are homotopic). This also shows that $X$ is stably equivalent to $S^2\vee \R P^2$.)

share|improve this answer
add comment

Your Answer


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.