TITLE: Reference for push-pull formula in cohomology
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: I think if you can consider the vertical maps in your diagram as fibre bundles, then there exists a commutative square with the same horizontal maps, but the vertical ones replaced by transfers maps (or Umkehr maps). Taking homology gives the formula. For the property mentioned above, you may look at 
Mann, Benjamin M.; Miller, Edward Y.; Miller, Haynes R. $S^1$-equivariant function spaces and characteristic classes. Trans. Amer. Math. Soc. 295 (1986), no. 1, 233–256.