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I think the easiest way to understand the Bockstein spectral sequence is through the exact couple coming from the long exact sequence of cohomology associated to $0\to\mathbb Z\to\mathbb Z\to \mathbb Z … In particular the difference between the dimension of $H^n(X,\mathbb Z/2)$ and that of the $Sq^1$-cohomology is equal to the number of $\mathbb Z/2$-factors in $H^n(X,\mathbb Z)$ and $H^{n+1}(X,\mathbb …

a): An element of $C^\times$ can be thought of as a pair $(a,b)$ of elements of $C$ with $ab=1$. This gives a) by applying of existence extension to $a$ and $b$ and unicity to $ab$ and $1$. b): The r …

points and to the integral Lie algebra cohomology. … Note however that the group scheme cohomology is definitely distinct from the Lie algebra cohomology and the group cohomology already in the case of the additive group scheme: The $2$-cocycle $((x+y)^p-x …

I missed that the question concerned the singular Kummer surface (which I think historically was what was what was called the Kummer surface but our current fixation on non-singularity has changed tha …

The mod $2$ cohomology of $\mathrm{SO}_n$ is not an exterior algebra as soon as $n\geq3$ (there is a unique non-zero element in degree $1$ whose square is non-zero, think of the case $\mathrm{SO}_{3}$ … Another example is $K(\mathbb Z,2)$ (aka $\mathbb C\mathbb P^\infty$), its integral cohomology ring is a polynomial ring on a degree $2$-generator while its homology ring is the free divided power algebra …

The answer to the first question is almost always no, see Roos, Jan-Erik(S-STOC) Derived functors of inverse limits revisited. (English summary) J. London Math. Soc. (2) 73 (2006), no. 1, 65--83. . A …

I would claim that the splitting (and indeed the whole universal coefficient theorem) is not really a topological theorem. If we take the homological version one really works with the chain complex $C …

This induces zero in cohomology (for trivial reasons) but is not null homotopic. (This is easily seen by an explicit calculation. …

even without torsion one can exploit that certain cohomology classes are not divisible by some particular integer). … In algebraic topology torsion (and more general integral cohomology again versus rational cohomology) are enormously important for understanding the homotopy type of a space. …

This is completely analogous to the case of étale cohomology where the first Chern class takes value in $H^2_{et}(X,\mathbb Z_\ell(1))$, where $\mathbb Z_\ell(1)$ is the inverse limit of $\{\mu_{\ell^n … Similarly the $n$'th Chern class lies most naturally in cohomology of $(2\pi i)^n\mathbb Z=(2\pi i\mathbb Z)^{\otimes n}$ resp. $\mathbb Z_\ell(n):=(\mathbb Z_\ell(1))^{\otimes n}$. …

These are comments on Dmitri's answer. I don't think the surface example can work as all holomorphic forms on a compact surface are closed (a result due to Kodaira I believe). The Cartan formula $L_v …

As Allen noticed, for rational cohomology it is enough to compute $T$-equivariant cohomology and then take $\Sigma_m$-invariants (if this is to work also for integral cohomology a more careful analysis … is free over the cohomology ring of $T$. …