# rational homology [closed]

what's the isomorphism between $H_*(X;\mathbb Q)$ and $H_*(X;\mathbb Z)\otimes \mathbb Q$

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## closed as unclear what you're asking by mez, Iuʇǝƃɹɐʇoɹ, Claude Leibovici, Jean-Claude Arbaut, Eric StuckyDec 9 '14 at 8:00

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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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nice edit! I was wondering how to see what the map was –  Juan S May 1 '11 at 1:08
so you are saying that $H_*(X;-)$ is a functor from the category of abelian groups to itself, taking an abelian group $G$ to the abelian group $H_*(X;G)$ and a morphism $f:G\longrightarrow H$ to a morphism $f_*:H_*(X;G)\longrightarrow H_*(X;H)$ sending $\sum{g_ix_i}, g_i\in G$ to $\sum{f(g_i)x_i}$, in particular the homomorphism $H_*(X;\mathbb Z)\longrightarrow H_*(X;\mathbb Q)$ is induced from the inclusion $\mathbb Z \hookrightarrow \mathbb Q$ –  studento May 1 '11 at 8:27
@student: indeed, most of that is what «$H_*(X,\mathord-)$ is a functor» means. –  Mariano Suárez-Alvarez May 16 '11 at 3:27

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.

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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 $\mbox{Tor}(H_{n-1}(X;\mathbb{Z}),\mathbb{Q})$?

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