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I asked this question on Mathematics SE three days ago, but didn't get the answer.

$\require{AMScd}$Let $G, H, K$ be groups and suppose that we have a diagram $$\begin{CD} G @>f_1>> H\\ @Vg_1VV\\ K \end{CD} $$ where morphisms $f_1$ and $g_1$ induce isomorphism in homology. Can one builds a group $U$ with the properties

  1. The following square commutes: $$\begin{CD} G @>f_1>> H\\ @Vg_1VV @VVg_2V\\ K @>>f_2> U \end{CD} $$
  2. $g_2$ and $f_2$ induce isomorphism in homology.

I tried to prove that $U:=K*_G H$ will be such a group. I used Mayer-Vietoris sequence, but it doesn't seem to be working.

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  • $\begingroup$ The group $U=K*_GH$ does have the required property, and the Mayer-Vietoris sequence proves this. What did you think was the problem with proving this using Mayer-Vietoris? $\endgroup$
    – IJL
    Commented Jun 5, 2018 at 10:26

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