Suppose I have two families of groups $A_k$ and $B_k$ indexed by the natural numbers and suppose $B_k$ acts on $A_k$. Suppose there are groups homomorphisms $A_{k+1} \rtimes B_{k+1} \to A_k \rtimes B_k$ extending maps $A_{k+1} \to A_k$ and $B_{k+1} \to B_k$. Also suppose that $\lim_k H_i(A_k)=0$ for $i>0$. Can I conclude that $\lim_k H_i(A_k \rtimes B_k) \cong \lim_k H_i(B_k)$? I am happy to make any finiteness assumptions about these homology groups and their limits.
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$\begingroup$ I made I typo. I want $B_k$ to act on $A_k$ so I should have flipped $A_k$ and $B_k$ in all semidirect products. $\endgroup$– 2223Commented Jul 27, 2017 at 22:27
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No, because it is not even true for constant families: let $A$ be an acyclic group, so $H_i(A)=0$ for $i>0$, and $B$ be a group which $A$ acts on interestingly, e.g. $B= F(A)$ is the free group on the underlying set of $A$. Then $H_1(B \rtimes A) = \mathbb{Z}$ but $H_1(B) = \mathbb{Z}^A$.
answered Jul 27, 2017 at 19:47
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2$\begingroup$ Unless you really meant that $B_k$ acts on $A_k$ but just wrote the semidirect product the wrong way round. $\endgroup$ Commented Jul 27, 2017 at 21:11
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2$\begingroup$ I think I wrote the semidirect product the wrong way. I want the acyclic group (in the limit) to be acted upon. $\endgroup$– 2223Commented Jul 27, 2017 at 21:28