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Why does there exist a map $X\rightarrow K(H_i(X;\mathbb Q),i)$ corresponding by the universal coefficient theorem to $H_i(X;\mathbb Z)\rightarrow H_i(X;\mathbb Q)$ induced from the inclusion $\mathbb Z\rightarrow \mathbb Q$?

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To give a map (up to homotopy) $X \to K(H_i(X; \mathbb{Q}), i)$ corresponds to giving a cohomology class in degree $i$ of $X$ with $H_i(X; \mathbb{Q})$-coefficients (if $X$ is nice). The identity map $H_i(X; \mathbb{Q}) \to H_i(X; \mathbb{Q})$ leads by dualization to a natural cohomology class of $H^i(X; H_i(X; \mathbb{Q}))$. (Since we're over a field, the Ext term in the universal coefficient theorem doesn't do anything.)

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thanks akhil that's a wonderful clear answer!!! –  palio Jun 9 '11 at 15:46

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