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

Question

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.

Answer 1

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.

Answer 2

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.