An interesting thing happened the other day. I was computing the Stiefel-Whitney numbers for $\mathbb{C}P^2$ **connect sum** $\mathbb{C}P^2$ to show that it was a boundary of another manifold. Of course, one can calculate the signature, check that it is non-zero and conclude that it can't be the boundary of an *oriented* manifold. I decided it might be interesting to calculate the first and only Pontrjagin number to check that it doesn't vanish. I believe Hirzebruch's Signature Theorem can be used to show that it is 6, but I was interested in relating the Stiefel-Whitney classes to the Pontrjagin classes.


I believe one relation is 



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> $p\_i (\mathrm{mod} 2) \equiv w\_{2i}^2$ (pg. 181 Milnor-Stasheff)
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So I went ahead and did a silly thing. I took my first Chern classes of the original connect sum pieces say 3a and 3b, used the fact that the inclusion should restrict my 2nd second "Stiefel-Whitney Class" (scare quotes because we haven't reduced mod 2) on each piece to these two to get $w\_2(connect sum)=(3\bar{a},3\bar{b})$. I can use the intersection form to square this and get $3\bar{a}^2+3\bar{b}^2=6c$ since the top dimensional elements in a connect sum are identified. Evaluating this against the fundamental class gives us exactly the first Pontrjagin number! **This is false. Of course this is wrong because it should be 9+9=18 as pointed out below. This does away with my supposed miracle example. My Apologies!**


This brings me to a broader question, namely of defining Stiefel-Whitney Classes over the integers. This was hinted at in [Ilya Grigoriev's response to Solbap's question](https://mathoverflow.net/questions/13813/construction-of-the-stiefel-whitney-and-chern-classes) when he says 



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> On thing that confuses me: why are the pullbacks of the integer cohomology of the real Grassmanian never called characteristic classes?
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Of course the natural reason to restrict to $\mathbb{Z}/2$ coefficients is to get around orientability concerns. But it seems like if we restrict our orientation to orientable bundles we could use a construction analogous to those of the Chern classes where Milnor-Stasheff inductively declare the top class to be the Euler class, then look at the orthogonal complement bundle to the total space minus its zero section and continue. I suppose the induction might break down because the complex structure is being used, but I don't see where explicitly. If someone could tell me where the complex structure is being used directly, I'd appreciate it. Note the Euler class on odd dimensional fibers will be 2-torsion so this might produce interesting behavior in this proposed S-W class extension.


Another way of extending Stiefel-Whitney classes would be to use Steenrod squares. Bredon does use Steenrod powers with coefficient groups other than $\mathbb{Z}/2$ (generally $\mathbb{Z}/p$ $p\neq 2$), but this creates awkward constraints on the cohomology groups. Is this an obstruction to extending it to $\mathbb{Z}$ coefficients? It would be interesting to see what these two proposed extensions of S-W classes do and how they are related.



Stiefel-Whitney Classes over Integers?
*I'm grateful to Allen Hatcher, who pointed out that this answer was incorrect. My apologies to readers and upvoters. 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 $\mathbb{Z}/(p^l)$ for all primes $p$ and natural numbers $l$, then they induce the same map on cohomology with $\mathbb{Z}$ coefficients. To see this, write $H^n(Y;\mathbb{Z})$ as a direct sum of $\mathbb{Z}^{r}$ and various primary summands $\mathbb{Z}/(p^k)$, and note that the summand $\mathbb{Z}/(p^k)$ restricts injectively to the mod $p^l$ cohomology when $l\geq k$. One can take only those $p^l$ such that there is $p^l$-torsion in $H^\ast(Y;\mathbb{Z})$. (I previously claimed that one could take $l=1$, which on reflection is pretty implausible, and is indeed wrong.)


We can try to apply this to $Y=BG$, for $G$ 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-primary torsion. That leaves the possibility that the mod 4 cohomology contains sharper information than the mod 2 cohomology. It does not, because, as Allen Hatcher has pointed out 
[in this recent answer](https://mathoverflow.net/questions/56932/what-characteristic-class-information-comes-from-the-2-torsion-of-hbsonz),
all the torsion is actually 2-torsion.


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.]


