Examples for non-naturality of universal coefficients theorem - MathOverflow

Examples for non-naturality of universal coefficients theorem

2011-03-29T04:11:48Z

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 $\mathbb{Z}/2$.

Answer by Sam Isaacson for Examples for non-naturality of universal coefficients theorem

2011-03-29T16:50:57Z

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

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

Answer by Tom Goodwillie for Examples for non-naturality of universal coefficients theorem

2011-03-31T02:09:35Z

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.

Answer by Charles Rezk for Examples for non-naturality of universal coefficients theorem

2011-03-31T22:56:01Z

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 $S^2\vee \R P^2$.)