# Tag Info

7

Here is a counterexample: let $$C_i = \begin{cases} \bigoplus_{n \in \mathbb{N}} \mathbb{Z} & i = 0 \\ 0 & i \neq 0 \end{cases}$$ with the zero differential. Then $H_0(C)=\bigoplus_{n \in \mathbb{N}} \mathbb{Z}$ has countably infinite rank; $H^0(C) = \prod_{n \in \mathbb{N}} \mathbb{Z}$ has a rank $2^{\aleph_0}$ by a theorem of Nöbeling. It's ...

4

The simplicial complex you describe is called the Tits building for $GL_n(k)$. When $k=\mathbb{F}_p$ it is homotopy equivalent to a wedge of $p^{n(n-1)/2}$ spheres of dimension $n-2$. I think that theorem is due to Quillen, and am pretty sure that volume II of Benson's Representations and Cohomology contains a proof. I don't know what happens for other ...

3

Related: curved $A_\infty$-algebras $\mathcal A:=(A_\bullet, d)$ are endowed with a pseudo differential $d_1:A_n\rightarrow A_{n+1}$ s.t. $$d_1\circ d_1(a)\pm d_2(a,d_0(1))\pm d_2(d_0(1),a)=0$$ for all $a\in \mathcal A$, denoting by $d_0(1)\in A_2$ the curvature of the algebra ( $1$ is the unit) and with $d_2$ the binary product in $\mathcal A$. Clearly, ...

2

Unless I'm being silly, the torsion subgroup is contained in the kernel of the map $H^{j+1}(X,\Bbb Z)\to H^{j+1}(X,\Bbb C)$, so you're done by exactness of the long exact sequence. (For reasonable spaces, e.g., manifolds or simplicial complexes, the sheaf cohomology and singular cohomology agree.)

2

Well, if those sets are pairwais disjoint, then everything splits as a direct sum: the $k$-forms $$\Omega^k(\cup_i U_i)\simeq\oplus_i\Omega^k(U_i)$$ is given by $\omega\mapsto (\omega|_{U_i})_i$ and the De Rham differential $d$ respects this splitting. So, both the kernel and image of $d$ respect this splitting as well.. Note, however, that for an infinite ...

2

As it came out in the comments, your doubt about cellular cohomology of $\mathbb{RP}^n$ not being isomorphic to singular cohomology was because you switched the even and odd cases in $\partial$, as seen on Hatcher (current online edition) p.144. In fact, cellular (co)homology is always isomorphic to singular (co)homology for CW complexes: Hatcher is again a ...

2

Use the universal coefficient theorem ($\mathbb{Z}_2$ is a PID): $$0 \to \operatorname{Ext}^1_{\mathbb{Z}_2}(H_{i-1}(M; \mathbb{Z}_2), \mathbb{Z}_2) \to H^i(M; \mathbb{Z}_2) \to \hom_{\mathbb{Z}_2}(H_i(M; \mathbb{Z}_2), \mathbb{Z}_2) \to 0$$ Since $\mathbb{Z}_2$ is a field, the $\operatorname{Ext}^1$ term is zero, and since $M$ is a closed (thanks Jack ...

2

Construct $K(G, n)$ by starting with $n$-cells and then attaching cells of higher dimension to kill the homotopy groups above dimension $n$; do the same for $K(G, m)$. The smash product will then have no cells below dimension $n + m$.

2

By the universal coefficient theorem $H^1(X)\cong\operatorname{Hom}(H_1(X),\Bbb Z)$ as $H_0$ is always free. Morphism groups to torsion-free groups are always torsion-free.

2

Are you familiar with the inflation restriction exact sequence? If $N \unlhd G$ and $G$ acts on module $A$, then $$0 → H^1(G/N, A^N) → H^1(G, A) → H^1(N, A)^{G/N} → H^2(G/N, A^N) →H^2(G, A)$$ is exact. Taking $N=H$ in your example, we have $A^N=0$, and $|N|$ and $|A|$ are coprime (I am assuming that $Z_{p^k}$ denotes a cyclic group of order $p^k$), so ...

1

Here are some books I've found helpful: Cohomology of Groups - Kenneth Brown. Very thorough, almost to a fault. It can take a while to actually get through because of how many details are included. If you really want to see the inner-workings of everything, work through this book and its exercises, although you may not develop any intuition in doing so. ...

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