Proof. Suppose that $(M_{i})_{i}$ is a collection of Landweber flat $\mathit{BP}^{\ast }$ -modules and let $n\geqslant m$ . By Proposition 2.4, it suffices to show that the top map in the commutative diagram

is injective. Since $P(m)^{\ast }/I_{m,n}$ is a finitely presented $P(m)^{\ast }$ -module, the vertical maps $\unicode[STIX]{x1D716}_{m}$ are isomorphisms. But, by assumption, the bottom map is injective, hence so is the top map.◻

Proof. We give the proof for $m=0$ and Morava $E$ -theory $E_{n}$ ; the argument and referenced results generalize easily to the $m>0$ case. First note that $E_{n}^{\ast }(X)$ is automatically complete with respect to the maximal ideal of $E_{n}^{\ast }$ , because the function spectrum $F(X,E_{n})$ is $K(n)$ -local. By [Reference Barthel and FranklandBF15, Proposition A.15], this means that $E_{n}^{\ast }(X)$ is flat as an $E_{n}^{\ast }$ -module if and only if it is pro-free, i.e. if it is the completion of some free $E_{n}^{\ast }$ -module. This reduces the claim to [Reference Hovey and StricklandHS99b, Proposition 2.5].◻

Proof. The base change formula for both $E(m,n)$ and $\hat{E}(m,n)$ is contained in the aforementioned references; see for example [Reference StricklandStr99, Corollary 8.24]. To show the second part of the claim, observe that, by [Reference WilsonWil99, Proposition 2.1], the assumption implies that the ideal $I_{m,n}$ acts regularly on $\hat{E}(m,n)^{\ast }(X)$ . Then the obvious extension of [Reference Hovey and StricklandHS99b, Theorem A.9] from $E_{n}$ to arbitrary $\hat{E}(m,n)$ gives flatness.◻

Proof. We prove this by induction on $n\geqslant m$ , the statement for $n=m$ being trivial. Now let $n\geqslant m$ and recall that Proposition 2.4 shows that $P(m)^{\ast }(X)$ being Landweber flat implies that

$$\begin{eqnarray}v_{n}:P(n)^{\ast }\otimes _{P(m)^{\ast }}P(m)^{\ast }(X)\longrightarrow P(n)^{\ast }\otimes _{P(m)^{\ast }}P(m)^{\ast }(X)\end{eqnarray}$$

is injective for all $n\geqslant m$ . By construction of $P(n+1)$ , there is an exact triangle

(2.1)

where the dashed arrow is the connecting homomorphism of degree 1. Since $P(n)^{\ast }(X)\cong P(n)^{\ast }\otimes _{P(m)}P(m)^{\ast }(X)$ by the induction hypothesis, our assumption combined with (2.1) shows that $\unicode[STIX]{x1D70C}_{n}$ is surjective and that

$$\begin{eqnarray}\displaystyle P(n+1)^{\ast }(X) & \cong & \displaystyle \text{coker}(v_{n}:P(n)^{\ast }(X)\rightarrow P(n)^{\ast }(X))\nonumber\\ \displaystyle & \cong & \displaystyle \text{coker}(v_{n}:P(n)^{\ast }\otimes _{P(m)^{\ast }}P(m)^{\ast }(X)\rightarrow P(n)^{\ast }\otimes _{P(m)^{\ast }}P(m)^{\ast }(X))\nonumber\\ \displaystyle & \cong & \displaystyle P(n+1)^{\ast }\otimes _{P(m)^{\ast }}P(m)^{\ast }(X),\nonumber\end{eqnarray}$$

thereby proving the claim. ◻

Proof. Let $X$ be a space such that $P(m)^{\ast }(X)$ is Landweber flat. From the previous lemma we get

(2.3) $$\begin{eqnarray}P(n)^{\ast }(X)\cong P(n)^{\ast }\otimes _{P(m)^{\ast }}P(m)^{\ast }(X)\end{eqnarray}$$

for all $n\geqslant m$ . But Proposition 2.10 with $m=n$ shows that

$$\begin{eqnarray}K(n)^{\ast }(X)\cong K(n)^{\ast }\,\,\hat{\otimes }_{P(n)^{\ast }}\,P(n)^{\ast }(X),\end{eqnarray}$$

which together with (2.3) yields the claim.◻

Proof. We will prove the result for $m=0$ using the classical Landweber filtration theorem. The argument in the $m>0$ case uses Theorem 2.1, but is easier as $P(m)$ is already $p$ -complete. Let $F$ be the fiber of the map $\mathit{BP}_{\!p}\rightarrow D$ and let

$$\begin{eqnarray}F^{\prime }=\text{fib}(\mathit{BP}\overset{}{\longrightarrow }D);\end{eqnarray}$$

we need to check that $F\overset{}{\longrightarrow }\mathit{BP}_{\!p}$ is null. Since $\mathit{BP}$ has degreewise finitely generated homotopy groups, $\mathit{BP}_{\!p}=I^{2}\mathit{BP}$ . In order to show the claim, it thus suffices by Lemma 3.1 to check that $F\overset{}{\longrightarrow }\mathit{BP}_{\!p}$ is $f$ -phantom. There is a commutative diagram

with all rows and columns being cofiber sequences. Since $D$ is $p$ -complete, we get $C_{3}=0$ and hence that the top row in the diagram

is a cofiber sequence. If we can show that $F^{\prime }\overset{}{\longrightarrow }\mathit{BP}$ is $f$ -phantom, then the composite $F^{\prime }\overset{}{\longrightarrow }\mathit{BP}\overset{}{\longrightarrow }\mathit{BP}_{\!p}$ is also $f$ -phantom, and hence is null by Lemma 3.1, so the indicated factorization in the right triangle exists. Because $\text{cof}(\mathit{BP}\overset{}{\longrightarrow }\mathit{BP}_{\!p})$ is acyclic with respect to the mod- $p$ Moore spectrum $M(p)$ and $\mathit{BP}_{\!p}$ is $M(p)$ -local, the dashed map must be null; hence so is $F\overset{}{\longrightarrow }\mathit{BP}_{\!p}$ .

Therefore, Spanier–Whitehead duality reduces the claim to proving that if $X$ is a finite spectrum, the induced map

is injective. By the Landweber filtration theorem, the $\mathit{BP}_{\ast }\mathit{BP}$ -comodule $\mathit{BP}_{\ast }(X)$ admits a finite filtration by comodules $F_{i}$ with filtration quotients $F_{i+1}/F_{i}\cong \mathit{BP}_{\ast }/I_{n_{i}}$ (up to suspensions). Since $\mathit{BP}_{\ast }/I_{n_{i}}\overset{}{\longrightarrow }D_{\ast }/I_{n_{i}}$ is injective by assumption, Landweber exactness of $D$ and the snake lemma applied to the commutative diagram

gives the claim inductively. ◻

Proof. It suffices to check that $D_{I}=\prod _{n\in I}\hat{E}(m,n)$ satisfies the conditions of Theorem 3.2. The spectra $\hat{E}(m,n)$ are Landweber exact and $K(n)$ -local, hence $p$ -complete. By Lemma 2.5, $D_{I}$ is also Landweber flat, and because products of local objects are local, it is also $p$ -complete. Since the standard maps $P(m)\overset{}{\longrightarrow }\hat{E}(m,n)$ clearly make

injective for all $k\geqslant m$ , the claim follows.◻

Proof. Since $\mathit{BP}$ is connective, $\mathit{BP}_{\!p}$ is the Bousfield localization of $\mathit{BP}$ with respect to the mod- $p$ Moore spectrum $M(p)$ , and we get a fiber sequence

(3.2)

where $C_{M(p)}\mathit{BP}$ is rational and concentrated in odd degrees. On the one hand, if $X$ is a rationally acyclic spectrum, then $C_{M(p)}\mathit{BP}^{\ast }(X)=0$ . On the other hand, if $X$ is a space that is rationally equivalent to the sphere, then $C_{M(p)}\mathit{BP}^{\ast }(X)\cong C_{M(p)}\mathit{BP}^{\ast }(S^{0})$ is concentrated in odd degrees. Moreover, the connecting homomorphism $\unicode[STIX]{x1D6FF}$ fitting in the commutative diagram

is surjective. Therefore, in either case the long exact sequence associated to (3.2) yields isomorphisms

and a short exact sequence

This implies the claim about evenness. To see that $\mathit{BP}^{\ast }(X)$ is Landweber flat if and only if $\mathit{BP}_{p}^{\ast }(X)$ is Landweber flat, note that the finite type condition together with the Atiyah–Hirzebruch spectral sequence show that $\mathit{BP}^{\ast }(X)$ is finitely generated in each degree. By [Reference BousfieldBou79, Proposition 2.5], we thus get an isomorphism

so the claim follows from Proposition 2.4 by using flatness of $\mathbb{Z}_{p}$ and naturality.◻

Proof. Let $I$ be the infinite set of natural numbers $n$ such that $K(n)^{\ast }(X)$ is even and $n\geqslant m$ . By Lemma 2.7, the assumption implies that $\hat{E}(m,n)^{\ast }(X)$ is even and flat over $\hat{E}(m,n)^{\ast }$ . Since the $P(m)^{\ast }$ -module $\hat{E}(m,n)^{\ast }$ is Landweber flat, so is $\hat{E}(m,n)^{\ast }(X)$ for all $n\in I$ . It follows from Lemma 2.5 that $\prod _{n\in I}\hat{E}(m,n)^{\ast }(X)$ is Landweber flat as well. Since Landweber flat modules are clearly closed under retracts, the claim now follows from Corollary 3.4.◻

Proof. By Theorem 3.9 with $m=0$ , $P(0)^{\ast }(X)$ is even, and hence so is $K(m)^{\ast }(X)$ for all $m>0$ by Corollary 2.13.◻

Proof. If $P(m)^{\ast }(X)$ is even, then it is also Landweber flat as the connecting homomorphism in (2.1) must be zero for degree reasons. Therefore, Corollary 2.13 shows that $K(n)^{\ast }(X)$ is even for all $n\geqslant m+1$ , so Corollary 3.11 applies.◻

Proof. Fix some integer $m\geqslant 0$ . We will prove the surjectivity statement; the argument for injectivity is analogous. By assumption and Lemma 2.7, there is a commutative diagram

Since $K(n)^{\ast }f$ is surjective, Lemma 3.14 implies that $\hat{E}(m,n)^{\ast }f$ is surjective for $n\in I$ as well. Products in the category of modules are exact, so the bottom map in the commutative square is also surjective:

Retracts of surjective maps are surjective, hence Corollary 3.4 yields the claim.

Assertion (ii) follows formally from (i) as in the proof of [Reference Ravenel, Wilson and YagitaRWY98, Theorem 1.18]. ◻

Proof. Let $n>0$ be such that $K(n)_{\ast }(X)$ is even. Recall that the completed $E$ -homology of $X$ is defined as $E_{n,\ast }^{\vee }(X)=\unicode[STIX]{x1D70B}_{\ast }L_{K(n)}(E_{n}\wedge X)$ . The assumption on $X$ implies that $E_{n,\ast }^{\vee }(X)$ is flat and even by the homological version of Lemma 2.7 (see [Reference Hovey and StricklandHS99b, Proposition 8.4(f)]), and so the previous discussion gives an isomorphism

of flat and evenly concentrated $E_{n}^{\ast }$ -modules, using the equivalence $E_{n}\wedge \mathbb{P}X\simeq \mathbb{P}^{E_{n}}(E_{n}\wedge X)$ . It follows from Lemma 2.7 that $K(n)^{\ast }(\mathbb{P}X)$ is even as well, so Theorem 3.9 finishes the proof.◻

Proof. Let $S$ be the set of natural numbers $n$ such that $K(n)_{\ast }(X)$ is even and degreewise finite. The free commutative algebra functor decomposes into its degree- $d$ constituents,

$$\begin{eqnarray}\mathbb{P}(-)\simeq \mathop{\vee }_{d\geqslant 0}\mathbb{P}_{d}(-)\simeq \mathop{\vee }_{d\geqslant 0}(-)_{h\unicode[STIX]{x1D6F4}_{d}}^{\wedge d},\end{eqnarray}$$

so we obtain a natural commutative diagram

for any spectrum $X$ . Since $\mathbb{T}^{\mathit{BP}}$ is compatible with this decomposition, we can reduce to a fixed degree $d\geqslant 0$ .

There are natural isomorphisms

$$\begin{eqnarray}\displaystyle \mathbb{T}_{d}^{\mathit{BP}}\mathit{BP}_{\!p}^{\ast }(X) & = & \displaystyle \unicode[STIX]{x1D716}\mathop{\prod }_{n\in S}\mathbb{T}_{d}^{E_{n}}(E_{n}^{\ast }\,\hat{\otimes }_{\mathit{BP}_{\!p}^{\ast }}\mathit{BP}_{\!p}^{\ast }(X))\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D716}\mathop{\prod }_{n\in S}\mathbb{T}_{d}^{E_{n}}(E_{n}^{\ast }(X))\quad \text{by Proposition 2.10}\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D716}\mathop{\prod }_{n\in S}\mathbb{T}_{d}^{E_{n}}\unicode[STIX]{x1D70B}_{\ast }E_{n}^{X}\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D716}\mathop{\prod }_{n\in S}\unicode[STIX]{x1D70B}_{\ast }L_{K(n)}\mathbb{P}_{d}^{E_{n}}E_{n}^{X}\quad \text{via}~\mathop{\prod }_{n\in S}\unicode[STIX]{x1D6FC}_{n}(E_{n}^{X}),\nonumber\end{eqnarray}$$

so it is enough to understand $\unicode[STIX]{x1D70B}_{\ast }L_{K(n)}\mathbb{P}_{d}^{E_{n}}E_{n}^{X}$ in terms of $E_{n}^{\ast }(\mathbb{P}_{d}X)$ . By [Reference Hovey and StricklandHS99b, Theorem 8.6], a spectrum $X$ has degreewise finite $K(n)_{\ast }(X)$ if and only if $X$ is dualizable in the $K(n)$ -local category; we write $D_{K(n)}$ for $K(n)$ -local duality. Therefore, we obtain

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70B}_{\ast }L_{K(n)}\mathbb{P}_{d}^{E_{n}}E_{n}^{X} & \cong & \displaystyle \unicode[STIX]{x1D70B}_{\ast }L_{K(n)}\mathbb{P}_{d}^{E_{n}}(E_{n}\wedge D_{K(n)}X)\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D70B}_{\ast }L_{K(n)}(E_{n}\wedge \mathbb{P}_{d}D_{K(n)}X)\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D70B}_{\ast }L_{K(n)}(E_{n}\wedge D_{K(n)}\mathbb{P}_{d}X)\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D70B}_{\ast }L_{K(n)}E_{n}^{\mathbb{P}_{d}X}\nonumber\\ \displaystyle & \cong & \displaystyle E_{n}^{\ast }(\mathbb{P}_{d}X).\nonumber\end{eqnarray}$$

Here, the third isomorphism uses the fact [Reference Greenlees and SadofskyGS96] that homotopy orbits agree with homotopy fixed points with respect to a finite group $G$ in the $K(n)$ -local category, i.e.  $Y_{hG}\overset{\simeq }{\longrightarrow }Y^{hG}$ $K(n)$ -locally. Moreover, the fourth isomorphism can be understood as follows: because $X$ is dualizable, $K(n)_{\ast }(X)$ is degreewise finite, and hence so is $K(n)_{\ast }(\mathbb{P}_{d}X)$ as $K(n)_{\ast }(B\unicode[STIX]{x1D6F4}_{d})$ is degreewise finite; using [Reference Hovey and StricklandHS99b, Theorem 8.6] again, we see that $\mathbb{P}_{d}X$ is also $K(n)$ -locally dualizable, giving the fourth isomorphism above.

Putting the pieces together, we obtain

$$\begin{eqnarray}\displaystyle \mathbb{T}_{d}^{\mathit{BP}}\mathit{BP}_{\!p}^{\ast }(X) & \cong & \displaystyle \unicode[STIX]{x1D716}\mathop{\prod }_{n\in S}\unicode[STIX]{x1D70B}_{\ast }L_{K(n)}\mathbb{P}_{d}^{E_{n}}E_{n}^{X}\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D716}\mathop{\prod }_{n\in S}E_{n}^{\ast }(\mathbb{P}_{d}X)\nonumber\\ \displaystyle & \cong & \displaystyle \mathit{BP}_{\!p}^{\ast }(\mathbb{P}_{d}X),\nonumber\end{eqnarray}$$

hence the claim. ◻

Proof. By assumption, Kuhn’s map $s_{n}(X)_{\ast }$ shows that $K(n)_{\ast }(\mathbb{P}X)$ is even, so Theorem 3.9 applies.◻

Proof. The implications $(1)\;\Longrightarrow \;(2)$ and $(3)\;\Longrightarrow \;(4)$ are trivial. Theorem 3.9 gives $(2)\;\Longrightarrow \;(3)$ , while $(4)\;\Longrightarrow \;(1)$ follows from Corollary 3.13.◻

Proof. For any formal group law there is a factorization

$$\begin{eqnarray}[p^{k}](x)=\underset{0\leqslant j\leqslant k}{\prod }\langle p^{j}\rangle (x).\end{eqnarray}$$

We will show that the factors are coprime in the rationalization; the Chinese remainder theorem will then imply the claim. We may write $\langle p\rangle (x)=p+f(x)$ for some power series $f(x)$ with $x|f(x)$ . This implies that

$$\begin{eqnarray}[p^{j}](x)=[p^{j-1}](x)(p+f([p^{j-1}](x)))\end{eqnarray}$$

and hence

(4.2) $$\begin{eqnarray}\langle p^{j}\rangle (x)=p+f([p^{j-1}](x)).\end{eqnarray}$$

But now if $t<j$ , we have $\langle p^{t}\rangle (x)|[p^{j-1}](x)$ by definition, so that

$$\begin{eqnarray}\langle p^{t}\rangle (x)|f([p^{j-1}](x)),\end{eqnarray}$$

and using (4.2) then gives

$$\begin{eqnarray}p=\langle p^{j}\rangle (x)-\langle p^{t}\rangle (x)\frac{f([p^{j-1}](x))}{\langle p^{t}\rangle (x)}.\Box\end{eqnarray}$$

Proof. We will prove the easiest case for clarity. Fix a height $n$ and let $k=1$ ; then with a coordinate this is the map

$$\begin{eqnarray}E_{n}^{\ast }[\![x]\!]/[p](x)\overset{}{\longrightarrow }E_{n}^{\ast }\times E_{n}^{\ast }[\![x]\!]/\langle p\rangle (x)\end{eqnarray}$$

sending $x\mapsto (0,x)$ and where $[p](x)$ now indicates the $p$ -series for the formal group law associated to $E_{n}$ . We may write down a basis for each side. For the left-hand side we have the basis $\{1,x,\ldots ,x^{p^{n}-1}\}$ , and for the right-hand side we can take the basis

$$\begin{eqnarray}\{(1,1),(0,1),(0,x),\ldots ,(0,x^{p^{n}-2})\}.\end{eqnarray}$$

It is clear that the basis elements $\{(1,1),(0,x),\ldots ,(0,x^{p^{n}-2})\}$ are hit under this map. Dividing by a power of $p$ is required to hit the basis element $(0,1)$ . But $\mathit{BP}$ provides a global element that hits $(0,1)$ at each height $n$ . It is the element (using the notation of the proof of the previous proposition)

$$\begin{eqnarray}-f(x)/p.\end{eqnarray}$$

Thus we see that we only need to divide by $p$ once in order to establish an isomorphism for all $n$ . For $k>1$ the proof is similar, but we need to divide by $p^{k}$ . For instance, when $k=2$ ,

$$\begin{eqnarray}\displaystyle -f([p](x)) & \mapsto & \displaystyle (0,0,p),\nonumber\\ \displaystyle -f(x) & \mapsto & \displaystyle (0,p,g(x)),\nonumber\end{eqnarray}$$

where $g(x)$ is some polynomial. This implies that

$$\begin{eqnarray}-\frac{f(x)}{p}+\frac{f([p](x))g(x)}{p^{2}}\mapsto (0,1,0).\end{eqnarray}$$

Finally, we see that the cokernel of the map

$$\begin{eqnarray}\underset{n>0}{\prod }E_{n}^{\ast }(B\mathbb{Z}/p^{k})\overset{}{\longrightarrow }\underset{n>0}{\prod }\,\underset{0\leqslant j\leqslant k}{\prod }E_{n}^{\ast }[\![x]\!]/\langle p^{j}\rangle (x)\end{eqnarray}$$

is all $p^{k}$ -torsion independently of the height. Thus the map is a rational isomorphism.◻

Proof. Let $\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{p})$ be the cokernel

We will show that $\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{p})$ is all $p$ -torsion with exponent $1$ , i.e. that it is annihilated by multiplication by  $p$ .

There is a stable splitting

$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{+}^{\infty }B\mathbb{Z}/p\simeq \unicode[STIX]{x1D6F4}_{+}^{\infty }B\unicode[STIX]{x1D6F4}_{p}\times Y\end{eqnarray}$$

where $Y$ is some spectrum. Applying $E$ -theory to this yields a commutative diagram

in which the first vertical map $i$ is split injective (by the stable splitting above). Since $E_{n}^{\ast }B\mathbb{Z}/p$ is free, the cokernel of $i$ is projective and hence $p$ -torsion free. Since $\mathbb{Q}\otimes \text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{p})=0$ , the kernel of the map $\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{p})\rightarrow \text{Cok}_{E_{n}}(\mathbb{Z}/p)$ must be torsion. Therefore the snake lemma implies that the kernel of the right vertical map $\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{p})\rightarrow \text{Cok}_{E_{n}}(\mathbb{Z}/p)$ must be a torsion subgroup of a torsion-free module and hence is zero. Thus $\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{p})$ is a subgroup of $\text{Cok}_{E_{n}}(\mathbb{Z}/p)$ . The claim now follows from Corollary 4.2.◻

Proof. The square in the proposition is the composite of two squares:

The right-hand square is an ‘arithmetic square’. It is a pullback square by applying [Reference Dwyer and GreenleesDG02, 4.12 and 4.13]. In [Reference Dwyer and GreenleesDG02] it is shown that the derived functors of localization and completion fit into a homotopy pullback square. The result we want is recovered by noticing that $S^{-1}R$ is free as an $S^{-1}R$ -module so that $H_{0}(S^{-1}R)=S^{-1}R$ , and that the derived completion and ordinary completion agree because $S^{-1}R$ is Noetherian.

The left-hand square is a pullback by direct computation. Let $r/s^{k}\in S^{-1}R$ and $r^{\prime }/p^{l}\in \mathbb{Q}\otimes R$ such that

$$\begin{eqnarray}\frac{r}{s^{k}}=\frac{r^{\prime }}{p^{l}}\in \mathbb{Q}\otimes S^{-1}R\end{eqnarray}$$

where $r,r^{\prime }\in R$ . Thus $s^{k}r^{\prime }=p^{l}r$ and

$$\begin{eqnarray}r\mapsto \frac{s^{k}r^{\prime }}{p^{l}}\in \mathbb{Q}\otimes R.\end{eqnarray}$$

But this implies that $s^{k}r^{\prime }/p^{l}\in R$ , since the map is an inclusion. Since $R$ is an integral domain and $s$ is, by definition, not in the ideal generated by  $p$ , we have

$$\begin{eqnarray}p^{l}|r^{\prime }\in R.\end{eqnarray}$$

Now $r^{\prime }/p^{l}$ is the pullback.◻

Proof. It suffices to set $k=l=1$ . We will use Lemma 4.4. There are $m_{1},m_{2}\in M$ such that $m_{1}/s\in S^{-1}M_{p}^{\wedge }$ maps to $x$ and $m_{2}/p\in p^{-1}M$ maps to $x$ . Now consider the commutative cube

The front and back faces are pullback squares. The elements $m_{1}/s$ and $m_{2}/p$ must agree in $p^{-1}(S^{-1}M_{p}^{\wedge })$ , and thus there is an element $m\in M$ that maps to both of these. By commutativity we must have $f(m)=x$ .◻

Proof. Let $K$ be the kernel

$$\begin{eqnarray}0\overset{}{\longrightarrow }K\overset{}{\longrightarrow }M\overset{}{\longrightarrow }\mathbb{Q}\otimes M=(\mathbb{Q}\otimes R)\otimes _{R}M.\end{eqnarray}$$

The $R$ -module $K$ is finitely generated since $R$ is Noetherian and $M$ is finitely generated. Choose generators $k_{1},\ldots ,k_{m}$ of $K$ ; then there exists an $n$ such that $nk_{i}=0$ for all $i$ . Now any element of $K$ is of the form $\sum r_{i}k_{i}$ , and this is also killed by  $n$ .◻

Proof. Consider the map of short exact sequences

The diagonal $\unicode[STIX]{x1D6E5}$ is a section of the projection onto a factor and the left vertical arrow is an equality, so $g$ is an injection. Thus there is an induced map $\text{Cok}(\unicode[STIX]{x1D6E5}f)\overset{r}{\longrightarrow }\text{Cok}(f)$ such that $rg=1$ . Since integer torsion is sent to integer torsion, this implies that the map induced by $g$ on integer torsion is a split injection. It is an isomorphism since a torsion element must be of the form $(m,m)\in M\times M$ with $(nm,nm)\in \unicode[STIX]{x1D6E5}fN$ , and so the kernel of $r$ is trivial.◻

Proof. Assume that $e(M)=n$ and let $[n]:M\overset{}{\longrightarrow }M$ be the multiplication-by- $n$ map. Consider the exact sequence

$$\begin{eqnarray}0\overset{}{\longrightarrow }K\overset{}{\longrightarrow }M\overset{[n]}{\longrightarrow }\text{im}[n]\overset{}{\longrightarrow }0,\end{eqnarray}$$

where $K$ is the torsion in $M$ . Base change to $T$ gives a short exact sequence and $T\otimes _{R}\text{im}[n]$ is necessarily torsion free; thus

$$\begin{eqnarray}e(T\otimes _{R}K)=e(T\otimes _{R}M)\leqslant n.\end{eqnarray}$$

If $T$ is faithfully flat, then $K\overset{}{\longrightarrow }T\otimes _{R}K$ is injective [Reference LamLam99, 4.74 and 4.83], so $e(T\otimes _{R}M)=e(M)$ .◻

Proof. By Lemma 4.5, the torsion in $N/\text{im}\,f$ is contained in the torsion of $(S^{-1}R)_{p}^{\wedge }\otimes _{R}N/\text{im}\,f$ , so

$$\begin{eqnarray}e(N/\text{im}\,f)\leqslant e((S^{-1}R)_{p}^{\wedge }\otimes _{R}N/\text{im}\,f).\end{eqnarray}$$

Also, $(S^{-1}R)_{p}^{\wedge }$ is a flat extension of $R$ , and this gives the reverse inequality. Since $T$ is faithfully flat, base change to $T$ is injective. Thus we have have equalities

$$\begin{eqnarray}e(N/\text{im}\,f)=e((S^{-1}R)_{p}^{\wedge }\otimes _{R}N/\text{im}\,f)=e(T\otimes _{R}(N/\text{im}\,f))\end{eqnarray}$$

by Proposition 4.9. ◻

Proof. The proof follows the proof of Proposition 6.2 in [Reference Barthel and StapletonBS16]. It is clear that the map is natural. To prove that it is an equivalence, the main idea is to note that the smash product on the left is $K(1)$ -local and then use the $K(1)$ -local self-duality of classifying spaces of finite groups.◻

Proof. This follows from the proof of Proposition 5.8 in [Reference Barthel and StapletonBS16], which relies on the important (and somewhat surprising) Proposition 3.13 that extends an argument of Hovey’s in [Reference HoveyHov04b]. Both module structures are faithfully flat because $K_{p}^{\ast }$ and $F_{1}^{\ast }$ are both complete local Noetherian.◻

Proof. This also follows the proof of Proposition 6.2 in [Reference Barthel and StapletonBS16], using Proposition 4.12 above. ◻

Proof. The map of $E_{\infty }$ -rings $E\overset{}{\longrightarrow }F$ and the transfer map $\unicode[STIX]{x1D6F4}_{+}^{\infty }BH\overset{}{\longrightarrow }\unicode[STIX]{x1D6F4}_{+}^{\infty }BG$ induce a commutative square of function spectra

and we can now base change the left-hand side to $F$ over $E$ .◻

Proof. The first isomorphism follows from the previous proposition and Proposition 4.11, and the second isomorphism follows from the previous proposition and Proposition 4.13. ◻

Proof. This is a basic application of HKR-character theory [Reference Hopkins, Kuhn and RavenelHKR00]. It is clear that the map is an isomorphism after base change to $C_{0}^{\ast }$ . The resulting ring is the ring of $C_{0}^{\ast }$ -valued functions on the set

$$\begin{eqnarray}\text{Sum}_{m}(\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n})=\underset{\unicode[STIX]{x1D706}\vdash m}{\coprod }\,\text{Sub}_{\unicode[STIX]{x1D706}\vdash m}(\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n}).\end{eqnarray}$$

But $C_{0}^{\ast }$ is a faithfully flat $p^{-1}E_{n}^{\ast }$ -algebra, so the map is a rational isomorphism.◻

Proof. Note that both the source and the target are finitely generated free $E_{n}^{\ast }$ -modules and the map is an injection. The cokernel $\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{m})$ is a torsion $E_{n}^{\ast }$ -module. We will repeatedly base change the short exact sequence

$$\begin{eqnarray}0\overset{}{\longrightarrow }E_{n}^{\ast }(B\unicode[STIX]{x1D6F4}_{m})\overset{}{\longrightarrow }\underset{\unicode[STIX]{x1D706}\vdash m}{\prod }\unicode[STIX]{x1D6E4}\,\text{Sub}_{\unicode[STIX]{x1D706}\vdash m}(\mathbb{G}_{E_{n}})\overset{}{\longrightarrow }\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{m})\overset{}{\longrightarrow }0\end{eqnarray}$$

to prove the result. Base change to $C_{1}^{\ast }$ (a flat $E_{n}^{\ast }$ -algebra) and (4.3) give the short exact sequence

$$\begin{eqnarray}\displaystyle \underset{[\unicode[STIX]{x1D6FD}]\in \text{Hom}(\mathbb{Z}_{p}^{n-1},\unicode[STIX]{x1D6F4}_{m})/\sim }{\prod }C_{1}^{\ast }\otimes _{L_{1}^{\ast }}L_{1}^{\ast }(BC(\text{im}\,\unicode[STIX]{x1D6FD})) & \overset{}{\longrightarrow } & \displaystyle \underset{\unicode[STIX]{x1D706}\vdash m}{\prod }\unicode[STIX]{x1D6E4}\,\text{Sub}_{\unicode[STIX]{x1D706}\vdash m}(\mathbb{G}_{C_{1}}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\nonumber\\ \displaystyle & \overset{}{\longrightarrow } & \displaystyle C_{1}^{\ast }\otimes _{L_{1}^{\ast }}L_{1}^{\ast }\otimes _{E_{n}^{\ast }}\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{m}).\nonumber\end{eqnarray}$$

Here we have used the fact that the scheme classifying subgroups behaves well under base change ([Reference StapletonSta15, Theorem 3.11] is particularly relevant). Corollary 4.10 implies that the exponent of the torsion cannot have changed. By faithful flatness (Proposition 4.9), we may remove the $C_{1}^{\ast }$ without changing the exponent of the torsion. Thus we arrive at the short exact sequence

(4.4) $$\begin{eqnarray}\underset{[\unicode[STIX]{x1D6FD}]}{\prod }L_{1}^{\ast }(BC(\text{im}\,\unicode[STIX]{x1D6FD}))\overset{}{\longrightarrow }\underset{\unicode[STIX]{x1D706}\vdash m}{\prod }\unicode[STIX]{x1D6E4}\,\text{Sub}_{\unicode[STIX]{x1D706}\vdash m}(\mathbb{G}_{L_{1}}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\overset{}{\longrightarrow }L_{1}^{\ast }\otimes _{E_{n}^{\ast }}\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{m}).\end{eqnarray}$$

Because the first and second terms are finitely generated and free, we may apply Corollary 4.10 to base change (4.4) to $F_{1}^{\ast }$ without changing the exponent of $L_{1}^{\ast }\otimes _{E_{n}^{\ast }}\text{Cok}_{E_{n}}(A)$ . We may then further base change to $\bar{F}_{1}^{\ast }$ (which is faithfully flat over $F_{1}^{\ast }$ by Proposition 4.12), obtaining

$$\begin{eqnarray}\underset{[\unicode[STIX]{x1D6FD}]}{\prod }\bar{F}_{1}^{\ast }(BC(\text{im}\,\unicode[STIX]{x1D6FD}))\overset{}{\longrightarrow }\underset{\unicode[STIX]{x1D706}\vdash m}{\prod }\unicode[STIX]{x1D6E4}\,\text{Sub}_{\unicode[STIX]{x1D706}\vdash m}(\mathbb{G}_{\bar{F}_{1}}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\overset{}{\longrightarrow }\bar{F}_{1}^{\ast }\otimes _{E_{n}^{\ast }}\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{m}).\end{eqnarray}$$

Now Proposition 4.13 gives the short exact sequence

$$\begin{eqnarray}\underset{[\unicode[STIX]{x1D6FD}]}{\prod }\bar{F}_{1}^{\ast }\otimes _{K_{p}^{\ast }}K_{p}^{\ast }(BC(\text{im}\,\unicode[STIX]{x1D6FD}))\overset{}{\longrightarrow }\underset{\unicode[STIX]{x1D706}\vdash m}{\prod }\unicode[STIX]{x1D6E4}\,\text{Sub}_{\unicode[STIX]{x1D706}\vdash m}(\mathbb{G}_{\bar{F}_{1}}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\overset{}{\longrightarrow }\bar{F}_{1}^{\ast }\otimes _{E_{n}^{\ast }}\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{m}).\end{eqnarray}$$

By the naturality of the equivalence, the first map in the sequence is induced by restriction maps. Let

$$\begin{eqnarray}M_{A,n}=\text{Cok}\biggl(\underset{[\unicode[STIX]{x1D6FD}]}{\prod }K_{p}^{\ast }(BC(\text{im}\,\unicode[STIX]{x1D6FD}))\overset{}{\longrightarrow }\underset{\unicode[STIX]{x1D706}\vdash m}{\prod }\unicode[STIX]{x1D6E4}\,\text{Sub}_{\unicode[STIX]{x1D706}\vdash m}(\mathbb{G}_{m}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\biggr);\end{eqnarray}$$

then

$$\begin{eqnarray}\bar{F}_{1}^{\ast }\otimes _{K_{p}^{\ast }}M_{A,n}\cong \bar{F}_{1}^{\ast }\otimes _{E_{n}^{\ast }}\text{Cok}_{E_{n}}(\unicode[STIX]{x1D6F4}_{m}).\end{eqnarray}$$

By Proposition 4.9, it suffices to show that the torsion in $M_{A,n}$ is independent of $n$ . The map

$$\begin{eqnarray}\underset{[\unicode[STIX]{x1D6FD}]}{\prod }K_{p}^{\ast }(BC(\text{im}\,\unicode[STIX]{x1D6FD}))\overset{}{\longrightarrow }\underset{\unicode[STIX]{x1D706}\vdash m}{\prod }\unicode[STIX]{x1D6E4}\,\text{Sub}_{\unicode[STIX]{x1D706}\vdash m}(\mathbb{G}_{m}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\end{eqnarray}$$

is the ring of functions on the canonical map

$$\begin{eqnarray}\underset{\unicode[STIX]{x1D706}\vdash m}{\coprod }\,\text{Sub}_{\unicode[STIX]{x1D706}\vdash m}(\mathbb{G}_{m}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\overset{f}{\longrightarrow }\text{Spec}(K_{p}^{\ast }({\mathcal{L}}^{n-1}B\unicode[STIX]{x1D6F4}_{m})).\end{eqnarray}$$

There is a commutative triangle

where

$$\begin{eqnarray}\text{Sum}_{{\leqslant}m}(\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})=\biggl\{\bigoplus _{i}H_{i}\;\biggm\vert\;H_{i}\subset \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1}~\text{and}~\mathop{\sum }_{i}|H_{i}|\leqslant m\biggr\}\end{eqnarray}$$

is the set of formal sums of subgroups in $\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1}$ with order no greater than $m$ . For each element $\unicode[STIX]{x1D6FC}\in \text{Sum}_{{\leqslant}m}(\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})$ , the map that lives over $\unicode[STIX]{x1D6FC}$ is a rational isomorphism (and an injection after applying $\unicode[STIX]{x1D6E4}(-)$ ). Thus it suffices to prove that, regardless of the choice of $n$ , there are only finitely many different maps that can appear as fibers. This is essentially a consequence of the fact that there are only finitely many groups that are centralizers of  $\unicode[STIX]{x1D6F4}_{m}$ .

To make this precise, note that there is an action of $\text{Aut}(\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})$ on the triangle. This action must send the map over $\unicode[STIX]{x1D6FC}\in \text{Sum}_{{\leqslant}m}(\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})$ to a map with the same exponent of the torsion in the cokernel. The action of $\text{Aut}(\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})$ is compatible with the inclusion of triangles induced by the inclusion of $\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1}\overset{}{\longrightarrow }\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n}$ as the first $n-1$ summands. For $n$ large enough, any sum of subgroups of size less than or equal to $m$ in $\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n}$ is isomorphic to a sum of subgroups in $\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1}$ under this action, and this finishes the proof.◻

Proof. Since the argument is similar to the proof of Theorem 4.17, we only point out the differences. We can again reduce to showing that the torsion in

$$\begin{eqnarray}M_{A,n}=\text{Cok}\biggl(\underset{\unicode[STIX]{x1D6FD}}{\prod }K_{p}^{\ast }(BA)\overset{}{\longrightarrow }\underset{H\subseteq A}{\prod }\unicode[STIX]{x1D6E4}\text{Level}(H^{\ast },\mathbb{G}_{m}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\biggr)\end{eqnarray}$$

has exponent independent of $n$ . To this end, note that the map

$$\begin{eqnarray}\underset{\unicode[STIX]{x1D6FD}}{\prod }K_{p}^{\ast }(BA)\overset{}{\longrightarrow }\underset{H\subseteq A}{\prod }\unicode[STIX]{x1D6E4}\text{Level}(H^{\ast },\mathbb{G}_{m}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\end{eqnarray}$$

is obtained by taking global sections of the canonical map

$$\begin{eqnarray}\underset{H\subset A}{\coprod }\text{Level}(H^{\ast },\mathbb{G}_{m}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\overset{f}{\longrightarrow }\text{Hom}(A^{\ast },\mathbb{G}_{m}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1}).\end{eqnarray}$$

It is a product of ring maps since we have the commutative triangle

and $\text{Hom}(A^{\ast },\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\cong \text{Hom}(\mathbb{Z}_{p}^{n-1},A)$ is a set and thus disconnects the horizontal map. The diagonal arrow sends a level structure $H^{\ast }{\hookrightarrow}\mathbb{G}_{m}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1}$ to the composite

$$\begin{eqnarray}A^{\ast }{\twoheadrightarrow}H^{\ast }{\hookrightarrow}\mathbb{G}_{m}\oplus \mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1}{\twoheadrightarrow}\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1}.\end{eqnarray}$$

Let $f_{\unicode[STIX]{x1D6FD}}$ be the fiber over $\unicode[STIX]{x1D6FD}$ . It is a map of schemes. It is clear that there is an action of $\text{Aut}(\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})$ on the triangle above. This implies that the fibers over maps $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FD}^{\prime }\in \text{Hom}(\mathbb{Z}_{p}^{n-1},A)$ with $\text{im}\,\unicode[STIX]{x1D6FD}=\text{im}\,\unicode[STIX]{x1D6FD}^{\prime }$ are isomorphic. Hence the exponent of the cokernel of $\unicode[STIX]{x1D6E4}(f_{\unicode[STIX]{x1D6FD}})$ equals the exponent of the cokernel of $\unicode[STIX]{x1D6E4}(f_{\unicode[STIX]{x1D6FD}^{\prime }})$ . Adding another $\mathbb{Q}_{p}/\mathbb{Z}_{p}^{}$ to the constant étale part does not affect this in the sense that the induced inclusion

$$\begin{eqnarray}\text{Hom}(A^{\ast },\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})\overset{}{\longrightarrow }\text{Hom}(A^{\ast },\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n})\end{eqnarray}$$

lifts to an inclusion of triangles. As $n$ increases above the number of generators of $A$ , $e(M_{A,n})$ remains constant. The fibers over all of the maps in $\text{Hom}(A^{\ast },\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n})$ are isomorphic to the fibers over maps that were already there in $\text{Hom}(A^{\ast },\mathbb{Q}_{p}/\mathbb{Z}_{p}^{n-1})$ . Thus the exponent of the torsion in $M_{A,n}$ is bounded independently of  $n$ .◻

Proof. Consider the composite

$$\begin{eqnarray}\underset{n}{\prod }E_{n}^{\ast }(BA)\overset{}{\longrightarrow }\underset{n}{\prod }\underset{H\subseteq A}{\prod }E_{n}^{\ast }(BH)/I\overset{}{\longrightarrow }\underset{n}{\prod }\underset{H\subseteq A}{\prod }\unicode[STIX]{x1D6E4}\text{Level}(H^{\ast },\mathbb{G}_{E_{n}}).\end{eqnarray}$$

By Proposition 4.19 the second map is a rational isomorphism, and by Theorem 4.18 the composite is a rational isomorphism. Thus the first map is a rational isomorphism. ◻

Proof. Let $H_{1},\ldots ,H_{m}\subseteq G$ be subgroups and consider the commutative diagram

where $I$ denotes the transfer ideal generated by $H_{1},\ldots ,H_{m}$ . The dashed arrows are the retracts which exist by Corollary 3.4, and the transfers make the left square commute by naturality and the fact that the splitting exists on the level of spectra. Therefore, the maps between the quotients exist and the resulting retract is compatible with the rest of the diagram. ◻

Proof. The argument is similar to the proof of Corollary 4.22. ◻

Proof. Because $G$ is good there is no integral torsion in $\mathit{BP}^{\ast }(BG)$ . This follows immediately from the injection

$$\begin{eqnarray}\mathit{BP}^{\ast }(BG){\hookrightarrow}\underset{n}{\prod }E_{n}^{\ast }(BG).\end{eqnarray}$$

Character theory for good groups [Reference Hopkins, Kuhn and RavenelHKR00] implies that the natural map

$$\begin{eqnarray}\underset{n}{\prod }E_{n}^{\ast }(BG)\overset{}{\longrightarrow }\mathop{\prod }_{n>0}\lim _{A\in {\mathcal{A}}({\mathcal{G}})}E_{n}^{\ast }(BA)\end{eqnarray}$$

is injective and, since limits of split injections are split injections, the canonical map

$$\begin{eqnarray}\lim _{A\in {\mathcal{A}}({\mathcal{G}})}\mathit{BP}^{\ast }(BA)\overset{}{\longrightarrow }\mathop{\prod }_{n>0}\lim _{A\in {\mathcal{A}}({\mathcal{G}})}E_{n}^{\ast }(BA)\end{eqnarray}$$

is injective.

We are interested in the commutative diagram

Since the left vertical arrow is injective and the second horizontal arrow on the bottom is injective, the natural map

$$\begin{eqnarray}\mathit{BP}^{\ast }(BG)\overset{}{\longrightarrow }\lim _{A\in {\mathcal{A}}({\mathcal{G}})}\mathit{BP}^{\ast }(BA)\end{eqnarray}$$

is injective. This implies that the top sequence of maps is short exact. Therefore it suffices to prove that $C_{\mathit{BP}}$ is torsion.

This means that it suffices to show that $C_{E_{n}}$ has exponent bounded independently of the height $n$ . By [Reference Hopkins, Kuhn and RavenelHKR00, Theorem A], the abelian group $C_{E_{n}}$ is torsion.

Note the map of short exact sequences

where $D_{E_{n}}$ is the cokernel. The first two vertical arrows are injections (the first is an equality), so the right vertical arrow is an injection. Therefore it suffices to show that the exponent of the torsion in $D_{E_{n}}$ is bounded independently of  $n$ .

To bound the torsion, we apply character theory to height 1. Since $C_{1}^{\ast }$ is a flat $E_{n}^{\ast }$ -module, we get a short exact sequence

$$\begin{eqnarray}C_{1}^{\ast }\otimes _{E_{n}^{\ast }}E_{n}^{\ast }(BG){\hookrightarrow}\underset{A\subset G}{\prod }C_{1}^{\ast }\otimes _{E_{n}^{\ast }}E_{n}^{\ast }(BA)\overset{}{\longrightarrow }C_{1}^{\ast }\otimes _{E_{n}^{\ast }}D_{E_{n}}.\end{eqnarray}$$

Now we can follow the argument in Theorem 4.18 to reduce to $p$ -adic $K$ -theory. This leads to considering the cokernel of the injection

$$\begin{eqnarray}\underset{[\unicode[STIX]{x1D6FC}]}{\prod }K_{p}^{\ast }(BC(\text{im}\,\unicode[STIX]{x1D6FC})){\hookrightarrow}\underset{A\subset G}{\prod }\underset{\unicode[STIX]{x1D6FD}}{\prod }K_{p}^{\ast }(BA).\end{eqnarray}$$

If we fix a conjugacy class $[\unicode[STIX]{x1D6FC}:\mathbb{Z}_{p}^{n-1}\rightarrow G]$ , then there is a map $K_{p}^{\ast }(BC(\text{im}\,\unicode[STIX]{x1D6FC}))\overset{}{\longrightarrow }K_{p}^{\ast }(BA)$ when the composite

$$\begin{eqnarray}\mathbb{Z}_{p}^{n-1}\overset{\unicode[STIX]{x1D6FD}}{\longrightarrow }A\subset G\end{eqnarray}$$

is conjugate to $\unicode[STIX]{x1D6FC}$ . In that case $A\subset C(\text{im}\,\unicode[STIX]{x1D6FC})$ . Let $i_{H}:A\subset G$ be the inclusion. The map out of the factor corresponding to $[\unicode[STIX]{x1D6FC}]$ is

$$\begin{eqnarray}K_{p}^{\ast }(BC(\text{im}\,\unicode[STIX]{x1D6FC}))\overset{}{\longrightarrow }\underset{A\subset G}{\prod }\biggl(\,\underset{\{\unicode[STIX]{x1D6FD}|[i_{H}\unicode[STIX]{x1D6FD}]=[\unicode[STIX]{x1D6FC}]\}}{\prod }K_{p}^{\ast }(BA)\biggr).\end{eqnarray}$$

This map is an injection, although it may not be the case that every abelian subgroup of $C(\text{im}\,\unicode[STIX]{x1D6FC})$ is represented in the codomain. Abelian subgroups of $G$ that are conjugate to abelian subgroups of $C(\text{im}\,\unicode[STIX]{x1D6FC})$ may appear as well. Also, subgroups may appear multiple times. However, Proposition 4.8 implies that repeat subgroups do not contribute to the exponent of the torsion in the cokernel. Finally, since there are only a finite number of abelian subgroups of $C(\text{im}\,\unicode[STIX]{x1D6FC})$ (or abelian subgroups of $G$ that may be conjugated into $C(\text{im}\,\unicode[STIX]{x1D6FC})$ ), the exponent of the torsion in the cokernel must be bounded by the maximal exponent of this finite number of options.◻