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Let $I$ be a directed set. Let consider the direct limit in the category of abelian groups. Suppose $A\cong \varinjlim_{i\in I} A_i$. Then, is it true that $A[2]\cong \varinjlim_i (A_i[2])$ ? Here, $[2]$ denotes the 2-part, that is, for abelian group $M$, $M[2]=\{x\in M\mid 2x=0\}$

Do you think does this hold in general ?

example 1. $\varinjlim_{i\in \Bbb{N}} \frac{1}{n}\Bbb{Z}\cong \Bbb{Q}$ and $\varinjlim_{i\in \Bbb{N}}(\dfrac{1}{n}\Bbb{Z})[2]\cong 0$ indeed holds.

But I don't have confident on counter examples. Thank you for very much for your help.

Poitou-Tate
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    Your example is not a direct limit but an inverse limit. – Captain Lama Jul 07 '24 at 13:11
  • Also, $(\mathbb{Z}/i\mathbb{Z})[2] = \mathbb{Z}/2\mathbb{Z}$ is only valid when $i$ is even. – Captain Lama Jul 07 '24 at 13:17
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    The maps of your limit on 2 torsion and all equal to $0$, so the limit is trivial as expected. Just think of the map $(\mathbb{Z}/4\mathbb{Z})[2] \to (\mathbb{Z}/2\mathbb{Z})[2]$. – Daniel5803 Jul 07 '24 at 13:22
  • And even for powers of $2$, you are completely neglecting what the transition maps are: the canonical reduction $\mathbb{Z}/4\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ induces the zero map $\mathbb{Z}/2\mathbb{Z}\simeq (\mathbb{Z}/4\mathbb{Z})[2]\to (\mathbb{Z}/2\mathbb{Z})[2]\simeq \mathbb{Z}/2\mathbb{Z}$. – Captain Lama Jul 07 '24 at 13:23

2 Answers2

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This is true if the colimit is filtered. Then $M[2] = \textrm{ker}(M \stackrel{2}{\to} M )$ is the kernel of a canonical map, and kernels commute with filtered colimits for abelian groups.

In other words: $$ \varinjlim M_i[2] = \varinjlim \textrm{ker}( M_i \stackrel{2}{\to } M_i) = \textrm{ker}( \varinjlim (M_i \stackrel{2}{\to } M_i)) = \textrm{ker}( M \stackrel{2}{\to } M) = M[2] $$

All equalities here are intended to be isomorphisms.

Andrea Marino
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  • Isn't $\varinjlim \textrm{ker}( M_i \stackrel{2}{\to } M_i) \cong \textrm{coker}( \varinjlim (M_i \stackrel{2}{\to } M_i))$ from universal property of direct limit ? – Poitou-Tate Jul 07 '24 at 13:45
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    No! Ker is a colimit, Coker is a limit. Unfortunately, no commutation rule holds in general between limits and colimits (think about them as product and sum, you do not have $(a+b)(c+d)=ab+cd$). However, when you have filtered colimits, it does commute with finite limits in cerain categories. See the discussion on filtered colimits on nLab for example – Andrea Marino Jul 07 '24 at 13:51
  • Changing inverse limit in linked page (https://math.stackexchange.com/questions/2013401/category-theory-limits-commute-with-kernels/2013415#comment10566972_2013415) into direct limit, I think formula at first my comment is indeed true and I cannot find the reason why your second '=' holds. – Poitou-Tate Jul 07 '24 at 14:22
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    Sorry, I made a typo in the above comment. Kernel is a limit and cokernel is a colimit. So you expect (as in the linked page) kernel to commute with limit, but not with colimit in general (except for the filtered case). Your formula cannot be right: if the diagram is made up of a single object, you would be saying that the kernel and cokernel of the map coincide. See https://ncatlab.org/nlab/show/filtered+colimit for the discussion about filtered colimit commuting with finite limits (in paritucalr kernels) of abelian groups, that justify my second passage. – Andrea Marino Jul 07 '24 at 15:30
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    Revisiting the universality diagram again, I realized that it indeed does not commute. Thanks to this, my research seems to be progressing. Thank you very much. – Poitou-Tate Jul 07 '24 at 16:41
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I think this should always be true. We have a natural map $\lim\limits_{\rightarrow}(A_i[2])\rightarrow(\lim\limits_{\rightarrow}A_i)[2]$ induced by the inclusions $A_i[2]\hookrightarrow A_i$. We can verify that this map is injective by the construction of direct limits. Now we check that it is surjective.

Consider an 2-torsion element $x$ in $\lim\limits_{\rightarrow}A_i$ represent by $a_i\in A_i$. Since $2a_i=0\in\lim\limits_{\rightarrow}A_i$, we know there exists an index $j\geq i$ such that $f_{ij}(2a_i)=0$ in $A_j$ where $f_{ij}:A_i\rightarrow A_j$ is the map in the direct system. Thus $f_{ij}(a_i)\in A_j[2]$, and hence the element represented by $f_{ij}(a_i)$ in $\lim\limits_{\rightarrow}(A_i[2])$ maps to $x$ under that natural map. So we have checked that the natural map at the beggining is an isomorphism.

LittleBear
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  • I'm not quite familiar with the terminology, but I guess "directed" might be a synonym of "filtered". So this answer coincides with Andrea Marino's above. – LittleBear Jul 07 '24 at 15:55