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I'm recently diving into the realm of homology and cohomology and encountered the universal coefficient theorems and concluded from it $H_i(X, k) = H_i(X,\mathbb{Z}) \otimes k$ for characteristic 0 fields $k$ ($X$ is a general top space). I'm also aware of the identity $H^i(X,k) = \text{Hom}_k(H_i(X,k),k)=H_i(X,k)^\vee$ (the dual space) for any field $k$ (no restriction on the characteristic). I'm curious about other similar identities:

  1. Does $H^i(X,k) = H^i(X,\mathbb{Z}) \otimes k$ hold for any char 0 field $k$? If yes, how to conclude it from the UCT? If no, when does this hold? When does it hold in greater generality, i.e. when is $H^i(X,G) = H^i(X,\mathbb{Z}) \otimes G$?
  2. It is $H_i(X,G) = H_i(X,\mathbb{Z}) \otimes G$, if $G$ or $H_{i-1}(X,\mathbb{Z})$ is torsion-free (as $\mathbb{Z}$-modules) by the UCT, because Tor vanishes in these cases. Is this right?
  3. Are there any other interesting identities relating the cohomology to the homology? An example is that the Betti-numbers are well-defined (either as $\mathbb{Q}$-dimension of the homology $H_i(X,\mathbb{Q})$ or as the $\mathbb{Q}$-dim of the cohomology $H^i(X,\mathbb{Q})$ (if all the (co)homology spaces are finitely generated, i.e. the (co)homology is of finite type).
  4. I've also read here The singular homology and cohomology of manifolds vanishes in high dimensions that the homology and cohomology of a connected compact $n$-manifold vanishes in degree $i>n$ and the homology and cohomology of a connected non-compact $n$-manifold vanishes in degree $i\geq n$. Am I confusing something or is this right? What about dropping the hypothesis of the connectedness of the manifold? Does it still hold?
Anzor Rubin
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1 Answers1

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  1. For $k$ a field of characteristic $0$, $H^i(X, k) \cong H^i(X) \otimes k$ holds if $H_i(X)$ and $H_{i-1}(X)$ are both finitely generated (by two applications of UCT) but not in general. Explicitly, take $i = 1, k = \mathbb{Q}$: then $H^1(X, \mathbb{Q}) \cong \text{Hom}(H_1(X), \mathbb{Q})$ and $H^1(X) \otimes \mathbb{Q} \cong \text{Hom}(H_1(X), \mathbb{Z}) \otimes \mathbb{Q}$ and it can happen that an abelian group admits nonzero maps to $\mathbb{Q}$ but doesn't admit nonzero maps to $\mathbb{Z}$, say $H_1(X) = \mathbb{Q}$.

  2. Yes, $H_i(X, G) \cong H_i(X) \otimes G$ if either $G$ or $H_{i-1}(X)$ is torsion-free.

  3. You can work through exactly what UCT implies in the case that all of the $H_i$ are finitely generated; then all of the $H^i$ are also finitely generated and the two determine each other (which is not true in general). If $X$ is a closed oriented manifold there is also Poincare duality.

  4. Yes, that's right. Connectedness doesn't matter, you just apply the result to each connected component.

Qiaochu Yuan
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