(Co-)Homology with different coefficients 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:

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*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 $H^i(X,G) = H^i(X,\mathbb{Z}) \otimes G$?

*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?

*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).

*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?

 A: *

*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 $H_1(X) = \mathbb{Q}$.


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


*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.


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