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

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

Najib Idrissi
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Lehi
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  • Yes, these statements follow from the universal coefficient theorems. – Aaron Mazel-Gee Jun 01 '11 at 16:05
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    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 01 '11 at 16:22

3 Answers3

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By the universal coefficients theorem, since a field is a PID, one has: $$H^n(X;\Bbbk) \cong \hom_\Bbbk(H_n(X;\Bbbk), \Bbbk) \oplus \operatorname{Ext}_\Bbbk^1(H_{n-1}(X;\Bbbk), \Bbbk).$$ But over a field all $\operatorname{Ext}$'s vanish, and thus: $$H^n(X;\Bbbk) \cong \hom_\Bbbk(H_n(X;\Bbbk),\Bbbk).$$

Now, if the homology groups with $\Bbbk$-coefficients are all finitely generated, then this means that $H^n(X;\Bbbk) \cong H_n(X;\Bbbk)$, because two vector spaces of the same dimension are isomorphic, and the dimension of the dual of a finite-dimensional vector space is the same. But in general, if $H_n(X;\Bbbk)$ is infinite dimensional, then depending on its (infinite) dimension, it may or may not be isomorphic to its dual.

For an explicit example, let $\Bbbk = \mathbb{Q}$, and let $X = \mathbb{N}$ be a countable discrete space. Then $H_0(X;\mathbb{Q}) = \bigoplus_{n \in \mathbb{N}} \mathbb{Q}$, while $H^0(X;\mathbb{Q}) \cong \prod_{n \in \mathbb{N}} \mathbb{Q}$, and the two are not isomorphic – the first has dimension $\aleph_0$, the second has dimension $2^{\aleph_0}$.

So to conclude, in general, you cannot say that $H_n(X;\Bbbk)$ is isomorphic to $H^n(X;\Bbbk)$; but if all the $H_n(X;\Bbbk)$ are finite dimensional, then it is true.

PS:

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

This is only valid if the homology groups are free and finitely generated. Again, same counterexample $X = \mathbb{N}$: $H_0(X;\mathbb{Z}) = \bigoplus_{n \in \mathbb{N}} \mathbb{Z}$ is free, but $H^0(X;\mathbb{Z}) = \prod_{n \in \mathbb{N}} \mathbb{Z} \not\cong H_0(X;\mathbb{Z})$.

(The other answers are not really explicit to answer the question in the title, so here you go.)

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

Thomas
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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 $H^i(X;A)\cong \text{Hom}(H_i(X;A),A)$.

wckronholm
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