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Suppose that we have (not necessarily injective) group homomorphisms $H \to G_1$ and $H \to G_2$, and we construct the pushout (i.e. amalgamated free product) $G_1 \sqcup_H G_2$. Suppose that we have a representation $V$ of the pushout group $G_1 \sqcup_H G_2$.

We can of course restrict the representation along the maps in the pushout square (I'll just denote the restricted reps by $V$ as well). In this case, is it possible to get a Mayer-Vietoris long exact sequence linking the group cohomology groups $\mathrm{H}^\bullet(H;V)$, $\mathrm{H}^\bullet(G_i;V)$, and $\mathrm{H}^\bullet(G_1 \sqcup_H G_2;V)$? If not, is there anything else we can say about the relationship between these cohomology groups?

(In case it matters, for my purposes $V$ will be a finite-dimensional real vector space.)

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    $\begingroup$ You always get a Mayer-Vietoris sequence if you compute the pushout in ∞-groups/E_1-groups: the square of classifying spaces is then a pushout square. Hence you get Mayer-Vietoris for the pushout in groups whenever the pushout in ∞-groups is again a group (i.e., 0-truncated). This is true if both maps are injective, but that's not necessary. $\endgroup$ Aug 25, 2022 at 16:17
  • $\begingroup$ @MarcHoyois Thanks! Do you happen to know of a reference where this is discussed? $\endgroup$
    – ಠ_ಠ
    Aug 25, 2022 at 21:25
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    $\begingroup$ It's a classical result that the classifying space functor B induces an equivalence between ∞-groups and pointed connected spaces (the inverse is the loop space). A modern reference is Theorem 5.2.6.10 in Lurie's Higher Algebra. It follows that the classifying space functor from ∞-groups to spaces commutes with pushouts, since pushouts preserve pointed connected spaces. $\endgroup$ Aug 26, 2022 at 5:50

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I had wondered about this myself once. When the maps $H\to G_i$ are injective, then Mayer-Vietoris in homology with trivial coefficients is discussed in chapter II, section 7 of Brown's Cohomology of groups. It should be fine for cohomology, and probably for more general coefficients (although I have not checked this carefully). For non injective maps, exercise 1 of that section of Brown says that everything works if $G_i$ and $H$ are replaced by their images in the pushout. However, if you don't do this, then Mayer-Vietoris won't work. To construct a counterexample, take $G_2=1$ and $H\to G_1$ surjective, then $G_1$ is the pushout. You won't get a Mayer-Vietoris exact sequence, unless the higher cohomology of $H$ is zero.

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