2 votes
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Are there examples where torsion subgroup of the first homology group is not $(\mathbb{Z}_2)^n$?

For $H_1 \approx \mathbb Z / n \mathbb Z$, take any manifold or CW complex space whose fundamental group is $\mathbb Z / n \mathbb Z$ (and apply the 1st Hurewicz theorem). You can use lens spaces, for ...
Lee Mosher's user avatar
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1 vote

Alternative universal coefficient theorem for cohomology?

You critically omit the condition that the homology of $X$ be of finite type listed under your link; without that assumption I doubt the formula is correct. With this assumption, on the other hand, ...
Ben Steffan's user avatar
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1 vote
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Cohomology ring of $S^2 \times S^2$ and $P2\#\bar{P2}$ are not isomorphic.

It suffices to show that there is no choice of basis for $H^1(\mathbb{P}^2 \# \overline{\mathbb{P}}^2)$, say $x' = ax + by$ and $y' = cx + dy$, such that $x'^2 = y'^2 = 0$. Since $x^2 + y^2 = xy = 0$ ...
ronno's user avatar
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1 vote
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Cohomology of $SO(3)$ bundle over $S^2$

That potentially nonzero differential lands in the $\mathbb{Z}/2\mathbb{Z}$, so if it's nontrivial, the kernel will still be $\mathbb{Z}$. This suggests the vanishing of $H^4$ of your total space can ...
Chris H's user avatar
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1 vote
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$H^i\left(BO_m ; \mathbb{Z}_2\right) \cong H^i\left(B O_n ; \mathbb{Z}_2\right)$ for $i \leq m$ and $i \leq n.$

As you state we have $H^*(BO_r; \mathbb{F}_2) \cong \mathbb{F}_2[w_1, \ldots, w_r]$, but you neglect to mention that $|w_i| = i$; with this the result is immediate since $\operatorname{rank}((\mathbb{...
Ben Steffan's user avatar
  • 2,947

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