I would like a precise reference for the following fact.

Assume that $$ \begin{array}{ccc} M\times_B N & \stackrel{f'}{\to} & N \newline \quad\downarrow g' & & \quad\downarrow g \newline M & \stackrel{f}{\to} & B \end{array} $$ is a Cartesian square of manifolds (meaning $f$ and $g$ are transverse, and $M\times_B N$ is their pullback, a submanifold of $M\times N$). Assume all manifolds are closed and oriented (although minimally we would require that $g$ is proper and oriented and $g'$ is endowed with the induced orientation). Then for any integral cohomology class $x\in H(N;\mathbb{Z})$ we have $$ f^*g_!(x) = g'_! f'^*(x)\in H(M;\mathbb{Z}). $$ Here $g_!:H^*(N;\mathbb{Z})\to H^{*+\dim(B)-\dim(N)}(B;\mathbb{Z})$ is the Umkehr map induced by $g$.

This is stated by Quillen for complex cobordism in his paper "Elementary proofs of some results in cobordism theory using Steenrod operations", and in fact it is immediate from his geometric interpretation of cobordism classes. It can also easily be deduced for integral cohomology, once one has some geometric interpretation of cohomology classes (such as Kreck's cobordism of stratifolds, or Buoncristiano-Rourke-Sandersons mock bundles), but I'm hoping it may be classical and appear in some older algebraic topology text.

proofbasically appears in that book, it's just never stated. He gets close when he's doing Lefschetz numbers, though... In any case, the general case is pretty easily reduced to the case of an embedding. Nowadays, you could also do a modern proof using Atiyah duality, I think. $\endgroup$ – Dylan Wilson Jul 22 '16 at 20:48