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Assume there are two directed systems $(A_i,f_{ij})$ and $(A'_i,f'_{ij})$ of groups over the directed set $(\mathbb{N},\le)$, such that for any $i,j$ there exists isomorphisms $\phi_{ij}:A_i\to A_i'$ and $\psi_{ij}:A_j\to A'_j$ such that $f'_{ij}\circ\phi_{ij}=\psi_{ij}\circ f_{ij}$. Is it true that $\varinjlim A_i$ is isomorphic to $\varinjlim A'_i$?

If it is true, is there some preferred isomorphism? What about direct limits in other categories and other directed sets?

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This is false.

Example. Let $A_i = A'_i = \mathbf Z^{(\mathbf N)}$ be an infinitely generated free abelian group. For $i < j$, let $f_{ij} \colon A_i \to A_j$ be the map that kills the first $j$ coordinates, and $f'_{ij} \colon A'_i \to A'_j$ the map that kills the even coordinates up to $2j$: \begin{align} f_i((a_j)_j) &= (0,\ldots,0,a_{j+1},\ldots),\\ f'_i((a_j)_j) &= (a_1,0,a_3,0,\ldots,a_{2j-1},0,a_{2j+1},a_{2j+2},\ldots). \end{align} It is clear that the $A_i$ and $A'_i$ are pairwise isomorphic (even in finite subsets). However, the colimits are not isomorphic: $$A = \underset{\longrightarrow}{\operatorname{colim}} A_i = 0,$$ since an element $(a_1,\ldots,a_j,0,0,\ldots)$ in some $A_i$ disappears in $A_j$. On the other hand, $$A' = \underset{\longrightarrow}{\operatorname{colim}} A'_i = \mathbf Z^{(2\mathbf N-1)}$$ consists of all finite sequences $(a_1,\ldots) \in \mathbf Z^{(\mathbf N)}$ supported on odd coordinates only. Indeed, the even coordinates all die by the same argument, and the odd ones never change in the colimit.

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  • $\begingroup$ Thanks a lot! If assume in addition that $A_i,A'_i$ are finitely generated abelian groups, will it be true? $\endgroup$ – ZZY May 29 at 18:33
  • $\begingroup$ I did try to think about this, but I was unable to come up with an example or a proof. The obvious proof is to extend to finite subsets of $\mathbf N$ first, and then to the limit. Both steps could go wrong (I think), and I see no reason to expect a positive answer. But please prove me wrong! $\endgroup$ – R. van Dobben de Bruyn May 29 at 18:52
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    $\begingroup$ Maybe I am missing something, but how is it obvious that these isomorphisms of $A_i$ and $A_i'$ commute with the $f_{ij}$ and $f_{ij}'$? OP does want the isomorphisms to be such that $f_{ij}' \circ \phi_{ij} = \psi_{ij} \circ f_{ij}$ (also to me it seems a bit odd that an isomorphism $A_i \to A_i'$ should also depend on $j$, so I am not sure if the way it is phrased now actually gives naturality). $\endgroup$ – Mark Kamsma May 30 at 16:58
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    $\begingroup$ @MarkKamsma: the dependence on $\phi$ and $\psi$ on both $i$ and $j$ is exactly the point of the question. If there were a single $\phi_i \colon A_i \stackrel\sim\longrightarrow A'_i$ commuting with all transition morphisms, these would trivially induce an isomorphism on the colimit. $\endgroup$ – R. van Dobben de Bruyn May 30 at 17:59
  • $\begingroup$ @R.vanDobbendeBruyn Oh right, now I see. I did indeed miss something, a slightly different construction for the isomorphisms than I first had in mind does work. $\endgroup$ – Mark Kamsma May 31 at 13:15

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