Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb Z\rightarrow \mathbb Q$?

share|cite|improve this question
up vote 1 down vote accepted

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.)

share|cite|improve this answer
thanks akhil that's a wonderful clear answer!!! – palio Jun 9 '11 at 15:46

Your Answer


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.