Formula for (total) chern classes in presence of a exact sequence with finite number of terms Suppose we have a connected smooth manifold $M$. And a short exact sequence of complex vector bundles on $M$. $$0 \rightarrow V_{1} \rightarrow … \rightarrow V_{n} \rightarrow 0 .$$
Then I recall that the product of $c(V_{i})$ for $i$ even is equal to product of  $c(V_{i})$ for $i$ odd. However I can only find the reference for $n=3$, i.e. short exact sequences. Does anyone know a reference for this? I flicked through Milnor Stasheff and didn't see it.

The idea is to split a long exact sequence into short exact sequences.  Let me use induction on $n$, the base case $n=3$ being known.  Suppose $n>3$ and the result is known for sequences of length $n-1$.  We can break an exact sequence $$0\to V_1\to \dots\to V_n\to 0$$ into two exact sequences $$0\to V_1\to \dots\to V_{n-2}\to W\to 0$$ and $$0\to W\to V_{n-1}\to V_n\to 0.$$  Let me assume $n$ is even; the odd case is similar.  By the induction hypothesis we get $$c(V_1)c(V_3)\dots c(V_{n-3})c(W)=c(V_2)c(V_4)\dots c(V_{n-2})$$ from the first sequence, and we also have $$c(V_{n-1})=c(W)c(V_n)$$ from the second sequence.  Multiplying these two equations and then cancelling $c(W)$ from both sides gives exactly what we want.  (We can cancel $c(W)$ since it is a unit in the ring $\prod_i H^i(M)$.  This is because the degree $0$ part of $c(W)$ is $1$, and so we can construct an inverse to $c(W)$ degree-by-degree.)
