# Confusion with universal coefficent theorem for cohomology

The universal coefficient theorem gives $$H^k(X,X';N)\cong Ext^1_R(H_{k-1}(X,X';R),N))\oplus Hom_R(H_k(X,X';R),N).$$

But now for any free abelian group $$A$$, we have $$Ext^1_{\mathbb Z}(A,\mathbb Z)\cong 0$$. If we take the homology to also have coefficients in $$\mathbb Z$$ then it seems that $$H_{k-1}(X,X';\mathbb Z)$$ is also a free group. This gives that $$H^k(X,X';\mathbb Z)\cong Hom_R(H_k(X,X';\mathbb Z),Z)$$.

I am confused as this seems to be too strong of a result. Is $$H_{k-1}(X,X';\mathbb Z)$$ not always a free group?

• $H_{k-1}(X, X';\mathbb{Z})$ is not always free, it could hypothetically be any abelian group. Commented Mar 3, 2020 at 14:53
• Definitely not. This is a somewhat "normal" mistake though. The chain complex on a space consists of free abelian groups, but once we take quotients, there is probably some torsion. Commented Mar 3, 2020 at 16:01
• okay, this seems like it was a good mistake to make. @William would you like to give a simple example as answer so I can accept it? Commented Mar 3, 2020 at 16:18

Here is an example of $$H_{k-1}(X, X';\mathbb{Z})$$ not being free: let $$X = \mathbb{R}P^{10}$$ and $$X'= \mathbb{R}P^5$$. Then $$H_i(X, X';\mathbb{Z}) \cong \mathbb{Z}/2$$ for $$i=7, 9$$.
More generally, if $$G$$ is any abelian group you could take $$X$$ to be a CW complex with $$H_{k-1}(X;\mathbb{Z})\cong G$$ and take $$X'$$ to be a subcomplex of dimension $$< k-2$$, then the relative homology will still be $$G$$.