Right exact group completion as a transfinite invariant of the homology equivalence

We consider a functor from the category of groups to itself $G\mapsto \mathbb Z_\infty G$thatwecallrightexact$\mathbb Z$-completion of a group. It is connected with the pronilpotent completion $\widehat{G}$ by the short exact sequence $1\to {\mathrm{}}^{}{M}_{n}G\to {\mathbb{Z}}_{\mathrm{\infty }}G\to \widehat{G}\to 1,$ where $M_{n}G$ is $n$-th Baer invariant of $G\mathrm{.}$ We prove that $\mathbb Z_\infty \pi_1(X)$isaninvariantofhomologicalequivalenceofaspace$X$. Moreover, we prove an analogue of Stallings’ theorem: if $G\to G^{\mathrm{\prime }}$ is a 2-connected group homomorphism, then ${\mathbb{Z}}_{\mathrm{\infty }}G\cong {\mathbb{Z}}_{\mathrm{\infty }}{G}^{\mathrm{\prime }}\mathrm{.}$ We give examples of $3$-manifolds $X,Y$ such that $ \hat{\pi_1(X)}\cong \hat{\pi_1( Y)}$but$\mathbb Z_\infty \pi_1(X)\not \cong \mathbb Z_\infty \pi_1(Y).$ We prove that for a finitely generated group $G$ we have $(\mathbb Z_\infty G)/ \gamma_\omega= \hat G.$Sothedifferencebetween$\hat G$and$\mathbb Z_\infty G$liesin$\gamma_\omega.$Thisallowsustotreat$\mathbb Z_\infty \pi_1(X)$asatransfiniteinvariantof$X.$ The advantage of our approach is that it can be used not only for $3$-manifolds but for arbitrary spaces.