# homology isomorphic to cohomology

is it true that when we compute homologies and cohomologies with coefficients in a field then homology and cohomology groups are isomorphic to eachother?

That is valid when homology groups are free with integer coefficients.

Thx

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Yes, these statements follow from the universal coefficient theorems. –  Aaron Mazel-Gee Jun 1 '11 at 16:05
No. What is true is that, as wckronholm says, the homology and cohomology are dual. This is not the same as saying they are isomorphic in infinite dimensions (e.g. consider $H_1$ and $H^1$ of a countable wedge of circles). –  Qiaochu Yuan Jun 1 '11 at 16:22

Given a space $X$ and an abelian group $A$, the Universal Coefficient Theorem for cohomology states that there is a natural short exact sequence $0\to \text{Ext}(H_{i-1}(X;\mathbb{Z}),A) \to H^i(X;A) \to \text{Hom}(H_i(X;\mathbb{Z}),A)\to 0$ and this sequence splits (but not naturally).

If $A$ is a field, then $\text{Ext}(H_{i-1}(X;\mathbb{Z}),A)=0$ and so $H^i(X;A)\cong \text{Hom}(H_i(X;A),A)$.

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For details, see pages 190-200 of Hatcher's book. –  wckronholm Jun 1 '11 at 16:13
does the ext become null if I take Z/2Z as a field? –  Lehi Jun 1 '11 at 16:21
@Lehi Yes! Ext is null for all fields, in particular for $\mathbb{Z}/2\mathbb{Z}$. –  wckronholm Jun 1 '11 at 16:26
In your last sentence: the $Ext$ is taken over abelian groups, so that $A$ is a field or not it quite irrelevant :) You really need to state a Universal coefficient theorem with your field $k$ in the place of $\mathbb Z$, and then do everything over $k$; in particular, the $Ext$s will then vanish. –  Mariano Suárez-Alvarez Jun 1 '11 at 16:59

The universal coefficient theorem is not needed in full generality. If $k$ is a field, then the functor $\text{Hom}(.,k)$ is exact. This gives a canonical isomorphism between cohomology and the vector space dual of homology.

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