Pulling Back product of Characteristic Classes

Question

I am reading Milnor & Stasheff's Characteristic Classes but I'm having difficulty understanding the following argument from the verification of Axiom 3 in Chapter 8.

$$w(\xi\times\xi') = w(\xi)\times w(\xi').$$ Now suppose that $\xi$ and $\xi'$, are bundles over a common base space $B$. Lifting both sides of this equation back to $B$ by means of the diagonal embedding $B\longrightarrow B\times B$, we obtain the required formula $$w(\xi\oplus\xi') = w(\xi)\cup w(\xi').$$

I understand that the Whitney sum is exactly the pullback of diagonal map hence the LHS of the conclusion follows But how do we get the RHS? Does this have something to do with the way product of cohomology classes is defined in terms of cup product?

This fact is again used in the proof of property 9.6 in the next chapter as follows.

$$e(\xi\times\xi') = (-1)^{mn}e(\xi)\times e(\xi');$$ where the sign can be ignored since the right side of this equation is an element of order two whenever $m$ or $n$ is odd.

Now suppose that $B=B'$. Pulling both sides of this equation back to $H^{m+n}(B;\mathbb{Z})$ by means of the diagonal embedding $B\longrightarrow B\times B$, we obtain the formula $e(\xi\oplus\xi') = e(\xi) \cup e(\xi')$ for the Euler class of a Whitney sum. $\qquad\square$

I think i am missing something obvious.

A little hint would do.

regards

Answer 1

Recall that for $x \in H^i(B_1)$ and $y \in H^j(B_2)$, the external product $a\times b \in H^{i+j}(B_1\times B_2)$ is defined as $p_1^*x\cup p_2^*y$ where $p_1 : B_1\times B_2 \to B_1$ and $p_2 : B_1\times B_2 \to B_2$ are the projection maps.

Now let $\Delta : B \to B\times B$ be the diagonal map $b \mapsto (b, b)$. Note that $(p_i\circ\Delta)(b) = p_i(\Delta(b)) = p_i(b, b) = b$ so $p_i\circ\Delta = \operatorname{id}_B$. Therefore $$\Delta^*(x\times y) = \Delta^*(p_1^*x\cup p_2^*y) = \Delta^*p_1^*x\cup\Delta^*p_2^*y = (p_1\circ\Delta)^*x\cup(p_2\circ\Delta)^*y = x\cup y.$$