Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Yes, these statements follow from the universal coefficient theorems. –  Aaron Mazel-Gee Jun 1 '11 at 16:05
3  
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
add comment

2 Answers

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

share|improve this answer
1  
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
4  
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.