rational homology what's the isomorphism  between $H_*(X;\mathbb Q)$ and $ H_*(X;\mathbb Z)\otimes \mathbb Q$
 A: 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 $\phi$.
A: 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.
A: 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 $\mbox{Tor}(H_{n-1}(X;\mathbb{Z}),\mathbb{Q})$?
