what's the isomorphism between $H_*(X;\mathbb Q)$ and $ H_*(X;\mathbb Z)\otimes \mathbb Q$
rational homology
3 Answers
There is a natural map of abelian groups $H_\bullet(X;\mathbb Z)\to H_\bullet(X;\mathbb Q)$, coming from the fact that $H_\bullet(X;\mathord-)$ is a functor, which we can tensor with $\mathbb Q$ over $\mathbb Z$, to get $\phi:H_\bullet(X;\mathbb Z)\otimes_{\mathbb Z}\mathbb Q\to H_\bullet(X;\mathbb Q)\otimes_{\mathbb Z}\mathbb Q.$ If you now notice that $H_\bullet(X;\mathbb Q)\otimes_{\mathbb Z}\mathbb Q$ is canonically isomorphic to $H_\bullet(X;\mathbb Q)$, because the latter is already a $\mathbb Q$-vector space, you see that the map you want is $\phi$.
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0@student: indeed, most of that is what «$H_*(X,\mathord-)$ is a functor» means. – 2011-05-16
The homology Universal Coefficient Theorem gives the short exact sequence $0 \to H_n(X,\mathbb Z) \otimes \mathbb Q \to H_n(X,\mathbb Q) \to \text{Tor}(H_{n-1}(X,\mathbb Z), \mathbb Q) \to 0.$
Loosely speaking, $\text{Tor}(A,B)$ measures the common torsion between $A$ and $B$. Since $\mathbb Q$ is torsion-free, the last term in the short exact sequence is trivial. This implies that the first map is an isomorphism.
From the universal coefficient theorem for homology we have the exact sequence $0 \to H_n(X;\mathbb{Z}) \otimes \mathbb{Q} \stackrel{\alpha}{\to} H_n(X; \mathbb{Q}) \to \mbox{Tor}(H_{n-1}(X;\mathbb{Z}),\mathbb{Q}) \to 0$ where $\alpha: (\mbox{cls} \ z) \otimes q \mapsto \mbox{cls}(z \otimes q)$
But what can you say about $\mbox{Tor}(H_{n-1}(X;\mathbb{Z}),\mathbb{Q})$?