Q:

Show that if $\widetilde{H}^n(X;\mathbb{Q})$ and $\widetilde{H}^n(X;\mathbb{Z_p})$ are zero, then $\widetilde{H}_n(X;\mathbb{Z})$ is zero.

Show that if $\widetilde{H}^n(X;\mathbb{Q})$ and  $\widetilde{H}^n(X;\mathbb{Z_p})$ are zero for all $n$ and all primes $p$, then $\widetilde{H}_n(X;\mathbb{Z})$ is zero for all $ n $.

My Try:
So I have the following corollary and wanted to use it here. 
Corollary: $\widetilde{H}_n(X;\mathbb{Z})=0$ iff $\widetilde{H}_n(X;\mathbb{Q})=0$ and $\widetilde{H}_n(X;\mathbb{Z_p})=0$ for all n and all primes p.
So now I wanted to show that

If $\widetilde{H}^n(X;\mathbb{Q})=0$ for all $n$ then $\widetilde{H}_n(X;\mathbb{Q})=0$ for all $n$ and 
If $\widetilde{H}^n(X;\mathbb{Z_p})=0$ for all $n$ and all prime $p$ then $\widetilde{H}_n(X;\mathbb{Z_p})=0$ for all $n$ and all prime $p$.

So suppose $\widetilde{H}^n(X;\mathbb{Q})=0$ for all $n$. Then $\ker \delta^n=$Im $ \delta^{n-1}$. Here $\delta ^{n-1}=\partial_n^*$ where $\partial_n $ is the map from $C_n(X)$ to $C_{n-1}(X)$. So,$\ker \partial_{n+1}^*=$Im $ \partial_n^*$. After that I was stuck. Now how can I show that $\ker \partial_n=$Im $ \partial_{n+1}$?

A:

To give an answer along the lines suggested by OP, one needs only remark that for a field $F$ we have a vector space isomorphism
$$Hom_F (H_n(X;F),F)\cong H^n(X;F)$$
Now  $H_n(X;F)$ is a vector space over $F$ and if $Hom_F (H_n(X;F),F)=0$ then 
$H_n(X;F)=0$.

