So the aim of this article will be to fill out the details of the computation of one can find in several books like “Adams’ Stable Homotopy and Generalised Homology” Lemma 2.52.5 page 39 or “Kochman’s Bordism, Stable Homotopy, and Adams Spectral Sequence Proposition 4.3.2 page 132. I feel the amount of details one usually found in these references are quite scarce. It could be a problem, since the topic is rather technical instead. Here is my attempt in filling the details.

Here is the Proposition :

Let be an oriented spectrum with orientation . Then:

- where is the inclusion map.
- where is the dual basis element of under the pairing .
- where is the dual basis element of under the pairing .

Before jumping to the proof, let us recall the concept of *oriented *spectrum.

**Def:** Let denote the inclusion map.** **A ring spectrum is called oriented spectrum if the following conditions are satisfied:

- is bounded below
- there is a class , called
*orientation class*, such that , the canonical generator of

Let me fix some notation: is the usual inclusion, and is the image of the orientation class in the *unreduced *second* *cohomology group of .

Clearly the Eilenberg-Maclane Spectrum is oriented for any ring . The *Thom spectrum * is oriented too (check on Kochman’s book, chapter 4.3)

I will prove only point 1 of the proposition for now. Maybe later I will update this article in order to include the proof of the other points, but really, there is nothing special *as soon you know how to prove the first. *

So let me prove the first point for you:

**Proof of (1):** The first step is to identify a certain element in the AHSS for with the orientation . Recall that

**Claim 1: **The element represents the orientation class .

**Proof of Claim 1: **Consider the AHSS for :

Since the edge homomorphism for the AHSS is always surjective, we have that the only possible non-zero differentials (the differentials of the 2nd page starting from the zero-th column to the second one) are trivial. This implies that the spectral sequence collapses. Since the second page is generated multiplicatively by and it is a free graded -module (i.e. the extension problem is trivial) the isomorphism maps ( and the other is the reduced cohomology group).

**Claim 2: **The AHSS for collapses at the second page and we have that represents the orientation class

**Proof of Claim 2:** Let us have a look at the AHSS for :

Since the AHSS is multiplicative and thanks to the ring structure of the second page, it’s enough to show that the element is an infinite cycle. In fact, one proceed inductively using the fact that any other element in the previous page (i.e. in the second page) is a -linear combination of powers of . Consider the inclusion . We know that is sent to by the fact that . Now recall that induces a map of spectral sequences. Since we know that AHSS for converges *a priori* to ( recall we assumed is required to be bounded below) there is an element in the stable page , for some which is a representative of . By **Claim 1**, we already know that the representative of lies in . Now suppose that the class of lives in , since the inclusion preserves the filtration, it would send our class to an element lying in , but since we already established that the image of lies in , it must be that and (Since ). By definition of the filtration for the cohomological AHSS, if this element lies in for in particular its representatives lie in meaning that when restricted to $\mathbb{C}P^1$ they are all zero (it is crucial here that the inclusion identifies the 2 skeleton of with ). Since we know that the restriction to of the orientation is a non-zero element, it must be and . This shows that we have to look for a representative for in . Consider the following diagram, where will denote the group in position p,q, page r of the AHSS for :

The lower map is an isomorphism since it is the map induced between the second singular cohomology groups of and . The diagram implies that the unique preimage of the representative of , which by **Claim 1** is identified , can only be the element in , and therefore it has to be an element of i.e. an infinite cycle. This readily implies that the AHSS collapses at the second page, since it is multiplicative: the second page is generated multiplicatively by $latex y \otimes \imath$, and we just showed that it is an infinite cycle, by an easy inductive reasoning we have that the differentials in every page must be zero since they are -linear derivations.

Since the stable page is a free graduated module, the extension problem is trivial. Therefore we have an isomorphism of -modules .

Using the fact that AHSS is multiplicative, and the ring structure on the stable page is the one one would expect to have, by compatibility of the two product structures (by axiom of multiplicative spectral sequence) and the above iso we have the claim. There is a nice discussion on Math.StackExchange about this proposition: I will update this answer accordingly