@BalarkaSen $\gamma^2$ not being the identity reminds me of an old question I asked here: Can torsion in the fundamental group happen for something embedded in $\mathbb{R}^3$?
suppose $\varphi : X \to \Bbb R^3$ is the embedding. clearly this gives an injection $\pi_1(X) \to \pi_1(\Bbb R^3)$, but the latter is zero so the former is zero, so everything embedded in $\Bbb R^3$ has trivial fundamental group, hence no torsion /s
Then if pi_2 is nonzero there's an embedded sphere representing it and your open manifold splits as a connected sum of things which embed in R^3. Since the free product of torsion-free groups is torsion-free (is this true?) you reduce to the case pi_2 = 0. Then X is a K(G,1) by the usual argument and is equivalent to a 2-complex so G can't have torsion.
global section is left adjoint to Spec... i.e. $\operatorname{Hom}(\Gamma (X,\mathcal O_X), A) = \operatorname{Hom}(X, \operatorname{Spec}(A))$? this feels wrong because something is contravariant @MatheinBoulomenos
@Leaky $\mathrm{Spec}$ and global sections (of ringed spaces, not of sheaves on a fixed base!) are both contravariant. Adjunctions for contravariant functors are defined as you would expect
@MikeMiller In this case you don't need to use Kurosh, I guess. You break up the submanifold of R^3 into connected sum of submanifolds with pi_2 = 0 in R^3 and each of those are finite-dimensional models of K(G_i, 1). The big submanifold has pi_1 = G_1 * ... * G_n and K(G_1 * ... * G_n, 1) = bigvee K(G_i, 1) which is a finite-dimensional model, so G_1 * ... * G_n is torsion-free.
@BalarkaSen yes Kurosh is sufficient: if A * B has torsion, it has a finite subgroup, which is the free product of a free group, a subgroup of A, and a subgroup of B. This can only happen if the free product has only one nontrivial factor
@LeakyNun there's no canonical way to do that for the reason mathei mentioned that you can have multiple scheme structures on a closed set - so usually you should be clear what scheme structure you're taking (most of the time its the reduced one)
I know how to write down the universal cover, and by cutting it into pieces I understand and how they fit together I get either an inductive MV description or an MV specseq depending on how you want to think about it
I'm guessing they first wrote it as $\displaystyle \prod_{p \le x} \sum_{n \ge 0}\frac{1}{p^n}$
But how does that weird index follow?
I'm only really interested in showing that the LHS > log(x).
There's a technique on the wiki page on zeta function but the method requires Re(s) > 1. This one diverges. And when we get to the harmonic series we easily get that this is > log(x).