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.

Question

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

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

2. 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}$?

Answer 1

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 $H_n(X;F)=0$.

Answer 2

As homology with $\mathbb Z$ coefficients consists of abelian groups, you can think of $\mathbb Q$-homology as measuring the free part and $\mathbb Z/p$-homology as measuring the $p$-torsion part. Hence we have: $X$ is an integer homology point iff $X$ is a rational homology point and a $\mathbb Z/p$-homology point forall $p$.

Next note that we always have by definition a natural map $$ H^i(X;R) \to Hom_R(H_i(X;R),R) $$

and by the universal coefficient theorem for $R$ a field we obtain that this is an isomorhism. In particular with field coefficients homology and cohomology are completely dual to each other (in linear algebra terms, by taking hom). In particular their dimensions as vector spaces coincide.

In particular we can extend:

$X$ is an integer homology point iff it is a rational cohomology point and a $\mathbb Z/p$-cohomology point for all $p$.