The Stiefel-Whitney classes of Cartesian product

I am reading the book of characteristic classes by Milnor-Stasheff, and I have a problem with the exercise 4-A: Show that the Stiefel-Whitney classes of a Cartesian product are given by $$w_k(\xi\times \eta)=\sum_{i=0}^k w_i(\xi)\times w_{k-i}(\eta).$$ Here is my solution: Let $B_1,B_2$ be base spaces of $\xi,\eta$ and $p_1,p_2$ be projections from $B_1\times B_2$ on $B_1, B_2$ respectively. By definition of the cross product, the right hand side of the above equality can be written as $$\sum_{i=0}^k p_1^{*}(w_i(\xi))\cup p_2^{*}(w_{k-i}(\eta)).$$ Also by the natuality axiom of Stiefel-Whitney classes, we have $p_1^{*}(w_i(\xi))=w_i(\xi \times \eta)$ and $p_2^{*}(w_{k-i}(\eta))=w_{k-i}(\xi \times \eta)$. So we can write $$\sum_{i=0}^k p_1^{*}(w_i(\xi))\cup p_2^{*}(w_{k-i}(\eta))=\sum_{i=0}^k w_i(\xi \times \eta)\cup w_{k-i}(\xi \times \eta),$$ which equals 0 if $k$ is odd (becasue we are working with coefficents in $\mathbb{Z}_2$).

I don't know where I am wrong. Somebody can help me?

• "Also by the natuality axiom of Stiefel-Whitney classes, we have $p_1^*(w_i(ξ))=w_i(ξ×η)$". Why should that be true?
– user64687
Oct 22 '13 at 11:02
• @AsalBeagDubh Because $p_1: B_1\times B_2\to B_1$ is covered by a bundle map from $\xi\times \eta$ to $\xi$. Oct 22 '13 at 11:32
• But that isn't how naturality works: it simply says that the SW classes of the pullback bundle are the pullbacks of the SW classes of the bundle. Think of it this way: suppose ξ is a trivial bundle of whatever rank. Then your claimed formula would say that $w_i(ξ×η)$ must be trivial for any bundle η. Does that seem plausible?
– user64687
Oct 22 '13 at 11:38
• @AsalBeagDubh Yes you were right. But that's the problem I am confused. Normally the naturality axiom I used here is in the book by Milnor-Stasheff, p39: maths.ed.ac.uk/~aar/papers/milnstas.pdf So can you explain why I was wrong? Oct 22 '13 at 11:54
• I think the confusion is what exactly is meant by a "bundle map". Look at the definition of "bundle map" on p.26 of the text, and Lemma 3.1 immediately after it. This says that if there is a bundle map $E \rightarrow F$, then $E$ is isomorphic to the pullback of $F$. So the naturality axiom then reduces to the statement I gave.
– user64687
Oct 22 '13 at 12:48

Also by the natuality axiom of Stiefel-Whitney classes, we have $p_1^{*}(w_i(\xi))=w_i(\xi \times \eta)$
Anyway, to solve the problem, you just need to write the Cartesian product bundle as $p_1^* \xi \oplus p_2^* \eta$ and then apply the Whitney sum formula.
• Let $\xi$ and $\eta$ have base spaces $B(\xi), B(\eta)$ respectively and total spaces $E(\xi), E(\eta)$, respectively. Now lift $\xi$ to subbundle of $\hat{\xi}\oplus \hat{\eta}$ on $B(\xi)\times B(\eta)$ by choosing $q\in B(\eta)$ so that $B(\hat{\xi}) = B(\xi)\times \{q\}$, and similarly for $\eta$: Let $B(\hat{\eta}) = \{p\} \times B(\eta)$. Now, $\hat{\xi}$ and $\hat{\eta}$ are each vector bundles over $B(\xi)\times B(\eta)$, and isomorphic to their summands in $\hat{\xi}\oplus \hat{\eta}$. Let $\pi^{-1}(x,y) = \pi_1^{-1}(x) \oplus \pi_2^{-1}(y)$. Trans fxns are $g_{ij}\oplus h_{kl}$, etc. Aug 6 '21 at 15:43