Let $f : X \rightarrow Y$ be a finite covering space. In [Quillen's paper](http://math1.unice.fr/~cazanave/Gdt/ImJ/Quillen.pdf) on the Adams conjecture in Topology, 1970, he gives the following argument for why 


$$f\_\* \psi^p = \psi^p f\_\* \in KO[p^{-1}]$$


Quillen's argument is that the map $f\_\*$ can be interpreted as a Gysin homomorphism, and "since the normal bundle of $f$ has a canonical trivialization, this Gysin homomorphism is essentially the composite of one suspension isomorphism and the inverse of another".


I'd like to flesh out this last part. How does one make it rigorous? 



Clarifying Quillen's comment on why Adams operations commute with transfer
Before I answer the question as stated, let me point out that you can also prove the claim as follows: the transfer is induced by a stable map $Y\_+\rightarrow X\_+$, and the pth Adams operation is a stable cohomology operation after inverting p, so the result is just naturality of cohomology operations.


Now for Quillen's argument. Embed $X$ into $S^n$ and so factor the map f as $X\rightarrow Y\times S^n \rightarrow Y$. If the normal bundle of the first map is, say, Spin oriented, then the Gysin map is defined as the composite: $KO(X) \rightarrow KO(Th(\nu)) \rightarrow KO(S^n \times Y) \rightarrow KO(Y)$. Here we use the Thom isomorphism, then the collapse map, then suspension isomorphism- I'm writing on a phone so I didn't put in the proper indices on the KO's or that some of those should be reduced K-theory.


Anyway, in this case the normal bundle in question is always trivial, and the Thom isomorphism for a trivial bundle is just the suspension isomorphism. This is what Quillen means by doing a suspension isomorphism and then undoing it.


The two arguments aren't unrelated of course, especially if you recall how the stable map yielding the transfer is constructed.


