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The Universal Coefficient Theorem for homology expresses the homology groups with coefficients in an arbitrary abelian group  in terms of the homology groups with coefficients in
 .
Theorem (Universal Coefficients for Homology)
Let  be a chain complex of free abelian groups, and let  any abelian group. Then there exists a split short exact sequence
As well,  and  are natural with respect to chain maps and homomorphisms of coefficient groups. The diagram splits naturally with respect to coefficient homomorphisms but not with respect to chain maps. Here, the functor
 is
 , the first left derived functor of
 .
We can define the map  as follows: Chose a cycle
![$ [u] \in H_n(K)$](http://images.planetmath.org:8080/cache/objects/4041/l2h/img12.png "$ [u] \in H_n(K)$") represented by  . Then
 is a cycle, so we set
![$ \alpha([u] \otimes x)$](http://images.planetmath.org:8080/cache/objects/4041/l2h/img15.png "$ \alpha([u] \otimes x)$") to be the homology class of
 . Of course, one must check that this is well defined, in that it does not depend on our representative for ![$ [u]$](http://images.planetmath.org:8080/cache/objects/4041/l2h/img17.png "$ [u]$") .
The universal coefficient theorem for cohomology expresses the cohomology groups of a complex in terms of its homology groups. More specifically we have the following
Theorem (Universal Coefficients for Cohomology)
Let  be a chain complex of free abelian groups, and let  be any abelian group. Then there exists a split short exact sequence
The homomorphisms  and  are natural with respect to coefficient homomorphisms and chain maps. The diagram splits naturally with respect to coefficient homomorphisms but not with respect to chain maps. Here
 is
 , the first right derived functor of
 .
The map  above is defined above in the following manner: Let
![$ [u] \in H^n($](http://images.planetmath.org:8080/cache/objects/4041/l2h/img27.png "$ [u] \in H^n($") Hom be represented by the cocycle
 Hom . For ![$ [x]$](http://images.planetmath.org:8080/cache/objects/4041/l2h/img31.png "$ [x]$") a cycle in  represented by  , we have
 . We therefore set
![$ \alpha([u])([x]) = u(x)$](http://images.planetmath.org:8080/cache/objects/4041/l2h/img35.png "$ \alpha([u])([x]) = u(x)$") . Again, it is necessary to check that this does not depend on the chosen representatives  and  .
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- W. Massey, Singular Homology Theory, Springer-Verlag, 1980
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![$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ 0 \ar[r] & H_n(K) \otimes_{\math... ...mathbb{Z}} G) \ar[r]^{\beta} & \mbox{Tor}(H_{n-1}(K),G) \ar[r] & 0.} } \end{xy}$](http://images.planetmath.org:8080/cache/objects/4041/l2h/img5.png "$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ 0 \ar[r] & H_n(K) \otimes_{\math... ...mathbb{Z}} G) \ar[r]^{\beta} & \mbox{Tor}(H_{n-1}(K),G) \ar[r] & 0.} } \end{xy}$")
![$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ 0 \ar[r] & \mbox{Ext}(H_{n-1}(K)... ...\mbox{Hom}(K,G)) \ar[r]^{\alpha} & \mbox{Hom}(H_n(K),G) \ar[r] & 0.} } \end{xy}$](http://images.planetmath.org:8080/cache/objects/4041/l2h/img20.png "$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ 0 \ar[r] & \mbox{Ext}(H_{n-1}(K)... ...\mbox{Hom}(K,G)) \ar[r]^{\alpha} & \mbox{Hom}(H_n(K),G) \ar[r] & 0.} } \end{xy}$")