Countable tower completion obstruction exact sequence

For $c=(c_m)\in L$, select $a_m\in c_m$ for every $m$ using [F2]. Compatibility says $a_m-a_{m+1}\in G_m$. Thus $b=(a_m-a_{m+1}{)}_m\in {\prod }_m{G}_m$, and set $\mathrm{\partial }(c)=[b]\in Q$. Another representative sequence has the form $a_m^{\mathrm{\prime }}=a_m+g_m$, with $g_m\in G_m$, and gives $b^{\mathrm{\prime }}=b+\mathrm{\Delta }(g)$. The class is therefore independent of every representative choice. Using the sequence $a_m+a_m^{\mathrm{\prime }}$ for the sum of two compatible families shows $\mathrm{\partial }(c+c^{\mathrm{\prime }})=\mathrm{\partial }(c)+\mathrm{\partial }(c^{\mathrm{\prime }})$. Hence $\mathrm{\partial }$ is a uniquely defined homomorphism, with no fixed section of any quotient included in its data.

The inclusion of ${\bigcap }_m{G}_m$ in $A$ is injective. The tuple $\eta (a)$ is compatible, and $\eta (a)=0$ exactly when $a\in G_m$ for every $m$. This proves exactness at the first two nonzero terms.

If $c=\eta (a)$, use the constant representative sequence $a_m=a$, whose difference is zero; thus $\mathrm{\partial }(c)=0$. Conversely, if $\mathrm{\partial }(c)=0$, a representative sequence from step 1.1 has difference $\mathrm{\Delta }(g)$ for some $g\in {\prod }_m{G}_m$. The elements $a_m-g_m$ then satisfy $a_m-g_m=a_{m+1}-g_{m+1}$ for every $m$, so all equal $a_0-g_0$. Their cosets are $c_m$, giving $c=\eta (a_0-g_0)$. This proves exactness at $L$ in both directions.

For any class $[b]\in Q$, take one tuple $b=(b_m)\in {\prod }_m{G}_m$ representing it. Define $a_0=0$ and $a_m=-{\sum }_{j<m}{b}_j$ for $m>0$. Then $a_m-a_{m+1}=b_m\in G_m$, so $(a_m+G_m{)}_m$ is compatible and maps to $[b]$. These finite sums require no choice, and a single existential representative of one quotient class requires no choice axiom. Thus $\mathrm{\partial }$ is surjective, proving the terminal exactness.

If $f:A\to A^{\mathrm{\prime }}$ preserves every subgroup, it sends a compatible tuple of cosets to a compatible tuple, and sends a representative sequence $a_m$ to $f(a_m)$. Its differences are $f(a_m-a_{m+1})$. The product map also commutes with $\mathrm{\Delta }$, so it induces the map on $Q$; the displayed sequence consequently commutes with $f$ at every term. If ${\bigcap }_m{G}_m=0$, step 1.2 makes $\eta$ injective, while steps 2.1–2.2 identify its cokernel with $Q$. Thus $\eta$ is an isomorphism exactly when $Q=0$. For $G_m=F_{-m}A$, the nonpositive indices are cofinal toward minus infinity: all other quotient components are uniquely determined by quotienting the component at zero. This limit is precisely the completion limit.

The zero group gives a zero sequence. If all $G_m=0$, then $L=A$ and $Q=0$. If all $G_m=A$, then $L=0$, the intersection is $A$, and step 2.2 shows $\mathrm{\Delta }$ is onto, so again $Q=0$. These constant cases show why the separatedness hypothesis is needed for the final equivalence with an isomorphism. No strict inclusions, finite generation or completeness of $A$ were assumed. The only countable selection was of the coset representatives in step 1.1.