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

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

share|cite|improve this question

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

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 $\phi$.

share|cite|improve this answer
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.

share|cite|improve this answer

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

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.