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For a group $G$, $H_2(G,\mathbb{Z})$ is also called the Schur multiplier of $G$. Among other things, if $G$ is perfect (ie $H_1=0$) then it is a term in the universal central extension $\widehat{G}$f …

I'm confident that hypotheses on the decision-theoretic properties of $\pi_1$ won't allow you to compute $\pi_2$. Below, I'll outline a construction that goes in this direction. Unfortunately, it doe …

I assume you are taking $M$ to be Riemannian and $\tilde{M}$ to be endowed with the induced metric? Let $\tilde{q}\in B_1(\tilde{p})$. We want to show that $\tilde{q}\in S$. Let $q=\pi(\tilde{q})$. …

There are some more-or-less equivalent conditions: Bass--Serre theory says that a group is freely indecomposable if and only if it acts on a tree with trivial edge stabilizers and no global fixed po …

The universal cover of a graph of groups is the Bass--Serre tree. This is described in Serre's book Trees, to which you refer. I don't have a copy to hand, so I can't give you the precise reference, …

An obvious point, but I hope worth making. The free group proof rests on the fact that graphs can be characterized locally. As tori can't be characterized locally by their topology, there's no hope …

As you implicitly point out, you might as well ask about conjugacy classes of subgroups of finite index in $F_2$. You might be interested in a famous paper of Marshall Hall Jr, in which he computed …

Theorem. Every finitely presentable group is the fundamental group of a closed 4-manifold. Sketch proof. Let $\langle a_1,\ldots,a_m\mid r_1,\ldots, r_n\rangle$ be a presentation. By van Kampen, th …

This question is very vague, but here are some thoughts to add to Mark's answer. First, note that any finitely presented group arises as the fundamental group of a closed manifold of dimension 4 (see …

This is certainly true, though one needs to be sufficiently careful about one's definition of Seifert fibred---it's important to allow fibres with a neighbourhood that looks like a fibred solid Klein …

This answer is really just intended to add some keywords to the discussion. If $G=\langle x_1,\ldots,x_m\mid r_1,\ldots,r_n\rangle$ then the set $\mathrm{Hom}(G,H)$ is naturally in bijection with the …

Regarding part 2 of your question - "In particular, how do we recognize infinite simple groups?" - I think the answer is that it depends on which infinite simple group you're looking at! Some famous …

According to Scott's paper "The geometries of 3-manifolds", the only closed manifolds that admit an $S^2 \times \mathbb{R}$ geometry are the two $S^2$ bundles over $S^1$, $P^2 \times S^1$ and $P^3 \# …

See pp. 449--457 of Peter Scott's article The geometries of 3-manifolds for a complete description of all 3-manifolds with finite fundamental group. The article is available on his website. There don …

They arise naturally all the time. The Galois correspondence tells us that a covering space is normal if and only if the corresponding subgroup of the fundamental group is normal. So whenever you hav …