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On the wikipedia page for Hodge cycles, it is stated that the universal coefficient theorem gives us a map $$H_k(M;\mathbb{Q})\to H_k(M;\mathbb{C})$$ But I don't see how. From what I know we would only obtain short exact sequences $$0\to H_k(M;\mathbb{Z})\otimes_\mathbb{Z}\mathbb{Q}\to H_k(M;\mathbb{Q})\to \text{Tor}(H_{k-1}(X;\mathbb{Z}),\mathbb{Q})\to 0$$ $$0\to H_k(M;\mathbb{Z})\otimes_\mathbb{Z}\mathbb{C}\to H_k(M;\mathbb{C})\to \text{Tor}(H_{k-1}(X;\mathbb{Z}),\mathbb{C})\to 0$$ And of course we have an inclusion $$H_k(M;\mathbb{Z})\otimes_\mathbb{Z}\mathbb{Q}\to H_k(M;\mathbb{Z})\otimes_\mathbb{Z}\mathbb{C}$$ but I do not see how any of this helps us in defining a map $H_k(M,\mathbb{Q})\to H_k(M,\mathbb{C})$. The only way in which I could see this work is if $$\text{Tor}(H_{k-1}(X;\mathbb{Z}),\mathbb{Q})=0$$ since then $$H_k(M;\mathbb{Z})\otimes_\mathbb{Z}\mathbb{Q}\cong H_k(M;\mathbb{Q})$$ and this will, when combined with the previously noted inclusion and the map $H_k(M;\mathbb{Z})\otimes_\mathbb{Z}\mathbb{C}\to H_k(M;\mathbb{C})$ give us the map we want. However, I'm not really experienced with $\text{Tor}$ and I do not see why $$\text{Tor}(H_{k-1}(X;\mathbb{Z}),\mathbb{Q})=0$$ Any help would be appreciated.

user2520938
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    There is no need to invoke the universal coefficient theorem. If $f : A \to B$ is any abelian group homomorphism, there's an induced natural transformation $H_k(-, A) \to H_k(-, B)$ given by applying $f$ to a singular chain in the obvious way. – Qiaochu Yuan May 13 '16 at 23:56
  • That is what I Imagined initially when I saw that we embed $H_k(M;\mathbb{Q})\to H_k(M;\mathbb{C})$, but the wikipedia page mentions the UCT explictely. But I think the two maps obtained in this way are the same right? The map that you were talking about and the map constructed using UCT – user2520938 May 14 '16 at 08:03

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