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Apr 2, 2016 at 14:59 comment added Daniel Grady @DenisNardin Thanks! I'll take a look.
Apr 2, 2016 at 14:55 comment added Denis Nardin Ok, now I see: the morphism you are studying is the boundary homomorphism that arises when $\beta$ is 0 in homology. I'm sorry for being pedantic, I was missing the key context that $H_*\beta$ is 0. It is probably related to theorem 1.7 in this paper math.jhu.edu/~wsw/papers2/math/10-boundary-J3M2WZ1-1975.pdf (they study a similar map for the Adams-Novikov spectral sequence but I believe the situations are very similar)
Apr 2, 2016 at 14:48 comment added Daniel Grady @DenisNardin I've fixed my original post. Is it clear now? Im looking at the map induced by the Bockstien on coefficients. It goes from the $E_2$ page of the AHSS for $E\wedge M\mathbb{R}/\mathbb{Z}$ to the $E_2$ page for $\Sigma E$.
Apr 2, 2016 at 14:41 history edited Daniel Grady CC BY-SA 3.0
deleted 110 characters in body
Apr 2, 2016 at 14:32 comment added Denis Nardin Then $\beta$ is not the Bockstein you're looking for. In fact it is still not clear what is the map you're looking for, even at the $E_2$-page level. Could you clarify what are the source and target of the map?
Apr 2, 2016 at 14:23 comment added Daniel Grady @DenisNardin Yes, $\beta$ induces a morphism of exact couples, but this is not the map I'm looking at. For example, take $K$ theory. Then $H^p(X;\pi_{-2q}(K\wedge M\mathbb{R}/\mathbb{Z}))\simeq H^p(X;\mathbb{R}/\mathbb{Z})$. $H^p(X;\pi_{-2q}(\Sigma K))\simeq H^p(X;0)\simeq 0$. So the map induced by $\beta$ is just $0$. What I really want is to look at is the Bockstien $\tilde{\beta}:H^p(X;\mathbb{R}/\mathbb{Z})\to H^{p+1}(X;\mathbb{Z})$.
Apr 2, 2016 at 13:38 comment added Denis Nardin I'm sorry if I'm being overly dense, but I don't understand. Are you looking for a map $H_*(X;(E\wedge M\mathbb{Z})_*)\to H_*(X;\Sigma E_*)$ that commutes with the differentials? If so I believe my argument (there's a map of exact couples) should provide that
Apr 2, 2016 at 13:35 comment added Daniel Grady @DenisNardin I just realized I wasn't very clear. I edited my post
Apr 2, 2016 at 13:34 history edited Daniel Grady CC BY-SA 3.0
added 333 characters in body
Apr 2, 2016 at 13:23 comment added Denis Nardin I'm confused. All morphisms of spectra induce a morphism of the corresponding AHSS, since you have a morphism of exact couples. Exactly where are you having problems?
Apr 2, 2016 at 13:06 history edited Sebastian Goette CC BY-SA 3.0
rectified some permutations of the letters "e" and "i"
Apr 2, 2016 at 11:46 history asked Daniel Grady CC BY-SA 3.0