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I'm interested in filling the details of the well known computation, via AHSS of $E^*(\mathbb{C}P^n)$; where $(E,x_E)$ is an oriented spectrum.

It is supposed to be easy, yet every proof I've read skip a (obvious?) detail which now I want to address. Let $i_n\colon \mathbb{C}P^n\to \mathbb{C}P^{\infty}$ be the usual inclusion.

Let us denote with $\mathbb{Z}[i_n^*x_H]/(i_n^*x_H)^{n+1}$ the cohomology ring $H^*(\mathbb{C}P^n)$. AHSS is multiplicative so everything boils down to prove that the element $$i_n^*x_H\otimes \imath \in E^{2,0}_2 \cong E^2((\mathbb{C}P^n)_2,(\mathbb{C}P^n)_1)\cong \widetilde{E}^2(\mathbb{C}P^1)$$ coincides with $i_n^*x_E$.

Once we know this the result is clear since the AHSS collapses and the stable page is a free $\pi_*E$-module.

In the additional hypothesis that $E$ is connective, then I can indeed manage to show that $i_n^*x_E=i_n^*x_H\otimes \imath$ (there is a little identification going on here): using l.e.s. of the pair $(\mathbb{C}P^n,\mathbb{C}P^{n-1})$ one gets $\widetilde{E}^2(\mathbb{C}P^n)\cong \widetilde{E}^2((\mathbb{C}P^n)_2)$ and using the iso $E^2(\mathbb{C}P^n)\cong \widetilde{E}^2(\mathbb{C}P^n)\oplus E^2(*)$ one has the claim.

But I want to drop this hypothesis. On every reference I found (Adams' Stable homotopy and generalised homology Lemma 2.5 page 39, Kochman's Bordism, Stable Homotopy and Adams Spectral Sequences prop 4.3.2 and in Ravenel Lemma 4.1.4) it is always given as obvious, without any assumption on $E$ other than being a ring spectrum and being oriented. I'm wondering now if I'm missing something really trivial

EDIt: This is an argument I sketched down and I'd love to have some feedback about it:

Let me fix some notation: $ i_n \colon \mathbb{C}P^n \to \mathbb{C}P^{\infty}$ is the usual inclusion, and $ y_E$ is the image of the orientation class in the unreduced second cohomology group of $\mathbb{C}P^{\infty}$.

The first step is to identify a certain element in the AHSS for $ \mathbb{C}P^1$ with the orientation $ i_1^* y_E$. Recall that $ H^*(\mathbb{C}P^n)\cong \mathbb{Z}[y]/(y^{n+1})$

Claim 1: The element $ y \otimes \imath \in E^{2,0}$ represents the orientation class $ i_1^*(y_E)$.

Proof of Claim 1: Consider the AHSS for $\mathbb{C}P^1$: enter image description here 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 $ y\otimes \imath \in E^{2,0}_2$ and it is a free graded $ \pi_*E$-module (i.e. the extension problem is trivial) the isomorphism $ E^2(\mathbb{C} P^1)\xrightarrow{\cong} E^{0,2}_2 \oplus E^{2,0}_2$ maps $ i_1^*(y_E) \mapsto y \otimes \imath$ ( $ E_2^{0,2} =E^2(\ast) $ and the other is the reduced cohomology group).

Claim 2: The AHSS for $ \mathbb{C} P^n$ collapses at the second page and we have that $ y \otimes \imath \in E_2^{2,0}$ represents the orientation class $ i_n^*(y_E)$

Proof of Claim 2: Let us have a look at the AHSS for $ \mathbb{C} P^n$: enter image description here

Since the AHSS is multiplicative and thanks to the ring structure of the second page, it's enough to show that the element $ y \otimes \imath \in E_2^{2,0}$ 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 $ \pi_*E$-linear combination of powers of $ y\otimes \imath$. Consider the inclusion $ i_1^n \colon \mathbb{C} P^1 \to \mathbb{C} P^n$. We know that $ i_n^*(y_E)$ is sent to $ i_1^*(y_E)$ by the fact that $ (i_1^n)^*i_n^*=i_1^*$. Now recall that $ i_1^n$ induces a map of spectral sequences. Since we know that AHSS for $ \mathbb{C} P^n$ converges a priori to $ E^*(\mathbb{C}P^n)$ ($ \pi_*E$ is required to be bounded below by Kochman and Adams in another chapter) there is an element in the stable page $ E_{\infty}^{p,q}$, for some $ p,q \in \mathbb{Z}$ which is a representative of $ i_n^*(y_E)$.

By Claim 1, we already know that the representative of $ i_1^*(y_E)$ lies in $$ E_{\infty}^{2,0}\cong F^2 E^2(\mathbb{C} P^1)/F^3 E^2(\mathbb{C} P^1)$$ Now suppose that the class of $$ i_n^*(y_E)\in F^p E^{p+q}(\mathbb{C} P^n)/F^{p+1} E^{p+q}(\mathbb{C} P^n)$$ since the inclusion preserves the filtration, it would send our class to an element lying in $ F^p E^{p+q}(\mathbb{C} P^1)/F^{p+1} E^{p+q}(\mathbb{C} P^1)$, but since we already established that the image of $$ i_n^*(y_E)\in F^2 E^2(\mathbb{C}P^1) / F^{3} E^2(\mathbb{C}P^1)$$ it must be that $ p\geq 2$ and $ q=0$ (Since $ F^pE^2(\mathbb{C} P^n)\subseteq F^{p-1}E^2(\mathbb{C} P^n)$). By definition of the filtration for the cohomological AHSS, if this element lies in $ F^p E^2(\mathbb{C}P^n) / F^{p+1} E^2(\mathbb{C}P^n)$ for $ p>2$ in particular its representatives lie in $$ F^2 E^2(\mathbb{C}P^n)$$ 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 $ \mathbb{C}P^n$ with $ \mathbb{C}P^1$). Since we know that the restriction to $ \mathbb{C}P^1$ of the orientation $ i_n^*(y_E)$ is a non-zero element, it must be $ p=2$ and $ q=0$. This shows that we have to look for a representative for $ i_n^*(y_E)$ in $E_{\infty}^{2,0}$. Consider the following diagram, where $ E_r'^{p,q}$ will denote the group in position p,q, page r of the AHSS for $ \mathbb{C}P^n$:

enter image description here

The lower map is an isomorphism since it is the map induced between the second singular cohomology groups of $ \mathbb{C} P^1$ and $ \mathbb{C} P^n$. The diagram implies that the unique preimage of the representative of $ i_1^*(y_E)\in E^2(\mathbb{C}P^1)$, which by Claim 1 is identified $ y \otimes \imath$, can only be the element $ y \otimes \imath$ in $ E_2'^{2,0}$.

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  • $\begingroup$ I am not sure that cohomology with $\mathbb{Z}$-coefficients is actually necessary to understand the proof. It is just directly known that $H^*(\mathbb{CP}^n; E^*) \cong E*[x]/x^{n+1}$ and here we can take as $x$ the pullback of an arbitrary $E^*$-module generator in $\tilde{H}^2(\mathbb{CP}^1, \mathbb{CP}^1;E^*) \cong \tilde{E}^*(\mathbb{CP}^1)$. $\endgroup$ Jun 12, 2016 at 13:50
  • $\begingroup$ @LennartMeier well yes, is it UCT right? I implicitly used it to write down my generator as $x_H\otimes \imath$ where $\imath$ is a $E^*$-module generator. So cohomology with integer coefficients is used to basically prove what you claim (I'm sure there are other ways, but I'm following what the refereces I wrote above suggest) $\endgroup$
    – Riccardo
    Jun 12, 2016 at 14:34
  • $\begingroup$ @Riccardo Can you clarify your definition of an oriented spectrum for me? The definition that I am used to is typically that you have a chosen element $x_E \in \tilde E^2(\Bbb{CP}^\infty)$ whose restriction to $\tilde E^2(\Bbb{CP}^1)$ is a particular fixed generator. $\endgroup$ Jun 12, 2016 at 18:42
  • $\begingroup$ @TylerLawson yes, together with $E$ is ring spectrum. (obvious but just to be clear once and for all). Kochman even asks for $\pi_*E$ being bounded below. $\endgroup$
    – Riccardo
    Jun 12, 2016 at 18:51
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    $\begingroup$ You might like to look at treatments of the equivariant case, where it is necessary to be a bit more careful and explicit about some details. One version is in Sections 4 and 5 of a memoir of mine at arxiv.org/pdf/math/0211058.pdf $\endgroup$ Jun 13, 2016 at 8:15

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