show/hide this revision's text 2 Corrected a significant mistake

Here's an exercise in universal coefficients

I'm grateful to Allen Hatcher, who pointed out that this answer was incorrect. My apologies to readers and the classification of f.gupvoters. abelian groups:I thought it more helpful to correct it than delete outright, but read critically.

If $X$ and $Y$ are cell complexes, finite in each degree, and two maps $f_0$ and $f_1\colon X\to Y$ induce the same map on cohomology with coefficients in $\mathbb{Q}$ and in $\mathbb{F}_p$ \mathbb{Z}/(p^l)$ for all primes $p$, p$ and natural numbers $l$, then they induce the same map on cohomology with $\mathbb{Z}$ coefficients. MoreoverTo see this, we may restrict attention to those primes $p$ such that $H^{\ast}(Y;\mathbb{Z})$ has write $p$-torsion.

Applying this to H^n(Y;\mathbb{Z})$ as a direct sum of $Y=BG$, for \mathbb{Z}^{r}$ and various primary summands $G$ a compact Lie group, you can see \mathbb{Z}/(p^k)$, and note that there's no more information in the integral characteristic classes of vector bundles with structure group summand $G$ than there is in \mathbb{Z}/(p^k)$ restricts injectively to the rational and mod $p$ characteristic classes for p^l$ cohomology when $l\geq k$. One can take only those $p$ p^l$ such that there is $H^{\ast}(BG)$ has p^l$-torsion in $p$-torsion. This H^\ast(Y;\mathbb{Z})$. (I previously claimed that one could take $l=1$, which on reflection is advantageous because, as Jason DeVito has remarkedpretty implausible, the structure of $H^*(BG;k)$ and is simpler when indeed wrong.)

We can try to apply this to $k$ is a field than when Y=BG$, for $k=\mathbb{Z}$.

WellG$ a compact Lie group. For example, $H^{\ast}(BU(n))$ is torsion-free (and Chern classes generate the integer cohomology), and so rational characteristic classes suffice. In $H^{\ast}(BO(n))$ and $H^{\ast}(BSO(n))$ there's only 2-torsion, and Pontryagin2-primary torsion. That leaves the possibility that the mod 4 cohomology contains sharper information than the mod 2 cohomology. It does not, S-W andbecause, as Allen Hatcher has pointed out in the latter casethis recent answer,Euler classes are all you needthe torsion is actually 2-torsion.In general, there will be a finite list of special primes for each $G$.

It's sometimes worthwhile to consider the integral Stiefel Whitney Stiefel-Whitney classes $W_{i+1}=\beta_2(w_i)\in H^{i+1}(X;\mathbb{Z})$, the Bockstein images of the usual ones. These classes are 2-torsion, and measure the obstruction to lifting $w_i$ to an integer class. For instance, an oriented vector bundle has a $\mathrm{Spin}^c$-structure iff $W_3=0$.
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Here's an exercise in universal coefficients and the classification of f.g. abelian groups: If $X$ and $Y$ are cell complexes, finite in each degree, and two maps $f_0$ and $f_1\colon X\to Y$ induce the same map on cohomology with coefficients in $\mathbb{Q}$ and in $\mathbb{F}_p$ for all primes $p$, then they induce the same map on cohomology with $\mathbb{Z}$ coefficients. Moreover, we may restrict attention to those primes $p$ such that $H^{\ast}(Y;\mathbb{Z})$ has $p$-torsion.

Applying this to $Y=BG$, for $G$ a compact Lie group, you can see that there's no more information in the integral characteristic classes of vector bundles with structure group $G$ than there is in the rational and mod $p$ characteristic classes for those $p$ such that $H^{\ast}(BG)$ has $p$-torsion. This is advantageous because, as Jason DeVito has remarked, the structure of $H^*(BG;k)$ is simpler when $k$ is a field than when $k=\mathbb{Z}$.

Well, $H^{\ast}(BU(n))$ is torsion-free (and Chern classes generate the integer cohomology). In $H^{\ast}(BO(n))$ and $H^{\ast}(BSO(n))$ there's only 2-torsion, and Pontryagin, S-W and, in the latter case, Euler classes are all you need. In general, there will be a finite list of special primes for each $G$.

It's sometimes worthwhile to consider the integral Stiefel Whitney classes $W_{i+1}=\beta_2(w_i)\in H^{i+1}(X;\mathbb{Z})$, the Bockstein images of the usual ones. These classes are 2-torsion, and measure the obstruction to lifting $w_i$ to an integer class. For instance, an oriented vector bundle has a $\mathrm{Spin}^c$-structure iff $W_3=0$.

[I'm sceptical of your example in $2\mathbb{CP}^2$. So far as I can see, $3a+3b$ squares to 18, not 6, and indeed, $p_1$ is not a square.]