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.

  • $\begingroup$ Yes, these statements follow from the universal coefficient theorems. $\endgroup$ – Aaron Mazel-Gee Jun 1 '11 at 16:05
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    $\begingroup$ 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). $\endgroup$ – Qiaochu Yuan Jun 1 '11 at 16:22

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.


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.)


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.

  • $\begingroup$ I believe you mean $Hom_k(.,k)$ is an exact functor, not $Hom_{\mathbb Z}(., k)$ functor, right? $\endgroup$ – Yilong Zhang Mar 6 '16 at 20:33
  • $\begingroup$ @YilongZhang yes, vector space duality! I edited my answer. $\endgroup$ – Thomas Mar 7 '16 at 13:19

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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    $\begingroup$ For details, see pages 190-200 of Hatcher's book. $\endgroup$ – wckronholm Jun 1 '11 at 16:13
  • $\begingroup$ does the ext become null if I take Z/2Z as a field? $\endgroup$ – Lehi Jun 1 '11 at 16:21
  • $\begingroup$ @Lehi Yes! Ext is null for all fields, in particular for $\mathbb{Z}/2\mathbb{Z}$. $\endgroup$ – wckronholm Jun 1 '11 at 16:26
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    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Jun 1 '11 at 16:59
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    $\begingroup$ Mariano is right. The statement as written is simply wrong. $\endgroup$ – KotelKanim May 24 '16 at 6:37

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