In Milnor/Stasheff Characteristic classes on page 180 there is a statement (inside a proof) that I don't fully understand.
Let $\tilde{G}_k$ be the oriented real grassmanian, $e$ its Euler class: there is an exact sequence $$ H^i(\tilde{G}_n)\stackrel{\smile e}{\longrightarrow} H^{i+n}(\tilde{G}_n) \stackrel{\lambda}{\longrightarrow} H^{i+n} (\tilde{G}_{n-1})\to H^{i+1}(\tilde{G}_n)\to\ldots $$ that comes from the Gysin sequence in which $\tilde{E}_0$---complement of the zero section of the canonical $n$-plane bundle over $\tilde{G}_n$---is replaced by $\tilde{G}_{n-1}$ via the cohomology isomorphism $f^*$ where $f: \tilde{E}_0\to\tilde{G}_{n-1}$ maps a "pointed $n$-plane" $(V,v)$ to the orthogonal complement of $v$ in $V$ with an induced orientation (probably).
The map $\lambda$ is a ring homomorphism as it can be expressed as $(f^*)^{-1}\circ \pi_0^*$ where $\pi_0: \tilde{E}_0\to \tilde{G}_n$ is the projection.
The books says that $\lambda$ maps the Pontryagin class $p_k(\tilde\gamma^n)$ to $p_k(\tilde\gamma^{n-1})$ and I cannot see why this is true.
After edit: Is this the right track of ideas? If yes, why is it co complicated?
(1) If $n$ is even, then I can see that the top class $p_{n/2}=e\smile e$ is mapped to zero by the exactness. Not sure how to prove the analogue for $n$ odd.
(2) The induction step (induction wrt "dimension of the bundle minus index of the class").
Let $\tilde\gamma^n$ be the canonical (real) bundle over $\tilde{G}_n$ and $\tilde\gamma_0^{n-1}$ the "orthogonal complement" bundle over $\tilde{E}_0=(\tilde\gamma^n)_0$. The map $f: \tilde{E}_0\to \tilde{G}_{n-1}$ is covered by a bundle map and by naturality of Pontryagin classes, we know that $f^*$ maps $p_k(\tilde\gamma^{n-1})$ to $p_k(\tilde\gamma_0^{n-1})$. So it is sufficient to prove that $\pi_0^*$ maps $p_k(\tilde\gamma^n)$ to $p_k(\tilde\gamma_0^{n-1})$. This is not obvious, as $\pi_0$ doesn't come from a bundle map.
Consider a bundle map $$ \begin{array}{ccc} A:=\tilde\gamma_0^{n-1}\otimes\mathbb{C} & \stackrel{\Phi}{\longrightarrow} & B:=((\tilde\gamma^n\otimes\mathbb{C})_0)^{2(n-1)} \\ \downarrow{q_1} & & \downarrow{q_2} \\ \tilde{E}_0=(\tilde\gamma^n)_0 & \stackrel{\phi}{\longrightarrow} & (\tilde\gamma^n\otimes\mathbb{C})_0 \end{array} $$ where $A$ consists of triples $(\tilde{V}, v, n_1\otimes 1+n_2\otimes i)$ such that $\tilde{V}$ is an oriented $n$-plane, $0\neq v\in\tilde{V}$ and $n_1, n_2\in (\tilde{V}\cap v^\perp)$. The map $\Phi$ maps such element to $$(\tilde{V}, v\otimes 1, n_1\otimes 1+n_2\otimes i)$$ and the projection $q_1$ maps it to $(\tilde{V}, v)$. By naturality, $\phi^*$ maps the Chern class $c_{2k}(B)$ to $c_{2k}(A)$.
Further, we have a commutative triangle of maps $$ \begin{array}{ccc} & \tilde{E}_0 &\\ {^{\pi_0}}\swarrow & & \searrow{^{\phi}}\\ \tilde{G}_n & \stackrel{\pi_1}{\longleftarrow} & (\tilde\gamma^n\otimes\mathbb{C})_0 \end{array} $$ where $\pi_1$ comes from the Gysin sequence of the bundle $\tilde\gamma^n\otimes\mathbb{C}$. If $k< n/2$, then both $\pi_0^*$ and $\pi_1^*$ induce cohomology isomorphism on the level of $H^{4k}$ (I hope). It follows that $\phi^*$ also induces cohomology isomorphism in dimension $4k$.
By the inductive definition of Chern classes, $\pi_1^*$ takes $c_{2k}(\tilde\gamma^n\otimes\mathbb{C})$ to $c_{2k}(B)$. We have already shown that $\phi^*$ takes $c_{2k}(B)$ to $c_{2k}(A)$. By the commutativity of the last diagram, $\pi_0^*$ takes $c_{2k}(\tilde\gamma^n\otimes\mathbb{C})$ to $c_{2k}(A)$ which is (up to sign) the Pontryagin class of $\tilde\gamma_0^{n-1}$.
Is this correct? (If yes, is this all so obvious that it is hidden in a half-sentence of Milnor's text?)