# Is it true that $A[2]\cong \varinjlim_i (A_i[2])$ if $A\cong \varinjlim_{i\in I} A_i$?

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.

• Your example is not a direct limit but an inverse limit. Commented Jul 7 at 13:11
• Also, $(\mathbb{Z}/i\mathbb{Z})[2] = \mathbb{Z}/2\mathbb{Z}$ is only valid when $i$ is even. Commented Jul 7 at 13:17
• 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 $(\mathbb{Z}/4\mathbb{Z})[2] \to (\mathbb{Z}/2\mathbb{Z})[2]$. Commented Jul 7 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 $\mathbb{Z}/2\mathbb{Z}\simeq (\mathbb{Z}/4\mathbb{Z})[2]\to (\mathbb{Z}/2\mathbb{Z})[2]\simeq \mathbb{Z}/2\mathbb{Z}$. Commented Jul 7 at 13:23

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.

• 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 ? Commented Jul 7 at 13:45
• 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 Commented Jul 7 at 13:51
• Changing inverse limit in linked page (math.stackexchange.com/questions/2013401/…) into direct limit, I think formula at first my comment is indeed true and I cannot find the reason why your second '=' holds. Commented Jul 7 at 14:22
• 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 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. Commented Jul 7 at 15:30
• 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. Commented Jul 7 at 16:41

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.

• 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. Commented Jul 7 at 15:55