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6

A non-orientable manifold has a connected double cover called the orientation double cover (see proposition 3.25 in Hatcher). Because it's a double cover, it induces an index-two subgroup of $\pi_1(M)$ (the subgroup of loops whose lifts are also loops). Quotienting out by this subgroup gives a nonzero homomorphism $\pi_1(M) \to \Bbb Z/2$, and hence (by the ...

2

You need only the following three facts: $H^1(X\sqcup Y) = H^1(X) \oplus H^1(Y)$, $H^1 (X\times \mathbb R) = H^1(X)$, and $H^1(\mathbb S^1) = \mathbb R$. The first and the third facts can be proved directly, while the second one (sometimes referred as the Poincare lemma) might be a bit harder to show. Everything are well discussed in the book ...

2

Take any embedded annulus $A : S^1 \times [0,1] \to M_K$ such that $A(S^1 \times 0)=m$. Since $A$ is embedded, $m'=A(S^1 \times 1)$ is disjoint from $m$. We may regard $A$ as a homotopy in $M_K$ from $m$ to $m'$, and crossing with $S^1$ we get a homotopy in $S^1 \times M_K$ from $F_1 = T_m = S^1 \times m$ to the disjoint torus $F_2 = T_{m'} = S^1 \times m'$. ...

2

One combinatorial way to do this is by using the Euler characteristic $$\chi(S^2) = V - E + F$$ as you suggested, where $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number faces in your simplicial complex. Since $\chi$ is a topological invariant, this number is the same for each triangulation, and also for each CW-complex. For a ...

1

I was thinking about this lying in bed this morning, and I believe that $I_G$ does contain elements $\zeta$ such that for all $E$, $\zeta_E$ is fixed by $g^*$ for all $g\in W_E$. Suppose $g\in W_E$ is not trivial. Then the following diagram commutes: $$\require{AMScd}\begin{CD} E @>{c_g}>> E \\ @VVV @VVV \\ G@>{c_g}>> G \end{CD}$$ ...

1

The fact that $$\operatorname{cor}\circ\operatorname{res}:H^n(G,A)\to H^n(G,A)$$ is multiplication by $(G:H)$ shows that the $p$-primary part of $H^n(G,A)$ embeds into $H^n(G_p,A)$, where $G_p$ is any Sylow $p$-subgroup. Since $H^n(G,A)$ is the direct sum of its $p$-primary parts, this shows that $H^n(G,A)$ is a subgroup of ...

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