I have a countable sequence of finite suspension spectra $X\_i$, whose $BP$-homology is a $BP\_\*(BP)$-comodule. Let's assume $BP\_\*(X\_i) = \Sigma^{d\_i} BP\_\* / (v\_0^{k\_0}, \dots v\_i^{k\_i}),$ for some $d\_n$ big enough that makes them being suspension spectra.


Then, if we let $X = \bigvee\_i X\_i$,


$$BP\_\*( X)= \bigoplus\_i BP\_\* / (v\_0^{k\_0}, \dots v\_i^{k\_i}).$$


Is that right? If yes/no, why?


Now, let's assume i have a spectrum $Y = \prod\_i X\_i$, which means


$$BP\_\* (Y) = \prod\_i BP\_\* / (v\_0^{k\_0}, \dots v\_i^{k\_i})$$ 


and the obvious map $$i: X \to Y$$


How would the cofiber $ C\_i$ of this map look like? More in detail, i would like $C\_i$ to have non torsion elements. Is this the case?


Thank you!



Homology of a limit of spectra + Cofiber
So far as I can tell, it depends.


The map $\oplus BP\_\*(X\_i) \to \prod BP\_\*(X\_i)$ is injective and the cokernel $M$ is $BP\_\*$ of the cofiber. In particular, for any $N$ we have an isomorphism
$$
M \cong \prod\_{i \geq 0} BP\_\*(X\_i) \Big/ \bigoplus\_{i \geq 0} BP\_\*(X\_i) \cong \prod\_{i \geq N} BP\_\*(X\_i) \Big/ \bigoplus\_{i \geq N} BP\_\*(X\_i)
$$
As a result, properties of the modules $BP\_\*(X\_i)$ that hold for all but finitely many exceptions tend to be inherited by the module $M$.


The first case is if the natural numbers $d\_n$ go to $\infty$ as $n$ goes to $\infty$. In the identities $BP\_\*(X) = \oplus BP\_\* (X\_i)$ and $BP\_\*(Y) = \prod BP\_\*(X\_i)$, the direct sum and product are taken in the category of *graded* abelian groups. If the $d\_n$ go to infinity, then for any fixed $k$ there are only finitely many $X\_i$ with $BP\_k(X\_i) \neq 0$. As a result the group $M$ is zero in grading $k$. (The same is true for homotopy groups, and so the cofiber of the map $X \to Y$ is contractible). If you're asking for the $X\_i$ to be suspension spectra, then I suspect that you're forced to have $d\_n \to \infty$ (but I don't have a proof).


The second case is when the integers $d\_n$ do not go to $\infty$. In this case, multiplication by $v\_j^{k\_j}$ acts by zero on all but finitely many $BP\_\*(X\_i)$, and so the same is true for $M$. Thus means that $M$ consists entirely of things that are torsion.




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What you might have had in mind was an "ultrafilter" type argument that did the exact opposite. If, instead, you have
$$
BP\_\*(X\_i) \cong \Sigma^{d\_i} BP\_\* / (v\_0^{k\_{0,i}}, v\_1^{k\_{1,i}}, \dots, v\_i^{k\_{i,i}})
$$
such that, for any fixed $n$, $k\_{n,i} \to \infty$, then you have the opposite case: for all but finitely many $i$, $BP\_m(X\_i)$ has no $v\_i^k$-torsion. As a result, the module $M$ has no torsion.




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(You probably already know this, but I feel obligated to mention that the identity $BP\_\*(\prod X\_i) \cong \prod BP\_\*(X\_i)$ isn't always true, and relies on the $X\_i$ being connective and $BP$ having ($p$-local) finite type.)


