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.

  • $\begingroup$ Bredon must have this. $\endgroup$
    – Thomas Rot
    Jul 22, 2016 at 9:19
  • $\begingroup$ @ThomasRot: Are you talking about his Sheaf cohomology book, or Topology and Geometry? I've briefly looked in both, but could take a closer look. $\endgroup$
    – Mark Grant
    Jul 22, 2016 at 9:21
  • $\begingroup$ I was thinking of Topology and geometry. I don't have the book in front of me so I could be mistaken. Sorry that my previous comment was so brief: Someone came in to my office while I was typing this and I forgot about it. $\endgroup$
    – Thomas Rot
    Jul 22, 2016 at 10:36
  • 2
    $\begingroup$ The statement when f and g are embeddings is in Bredon's Topology and Geometry (and many other books)... after looking through the sections a bit, I think the proof basically 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. $\endgroup$ Jul 22, 2016 at 20:48
  • $\begingroup$ @DylanWilson: Ah yes, I think I remember how to prove it now (factorize $g$ through an embedding into a trivial bundle over $B$, then use naturality of Thom and suspension isomorphisms). I'm a little surprised that there does not seem to be a good reference for this statement in full generality, but I'm happy to accept that it might not need one :) $\endgroup$
    – Mark Grant
    Jul 25, 2016 at 8:57

1 Answer 1


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.


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