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Let $p$ be a prime number and $X$ a topological space. Are the following equivalent?

(1) In the homology module $H_*(X;\mathbb{Z})$ there does not exist any element of order $p$.

(2) In the cohomology module $H^*(X;\mathbb{Z})$ there does not exist any element of order $p$.

Could you explain it? I want to use the Universal Coefficient Theorem but do not know how to use. We can impose extra conditions, such as finite $CW$-complexes, or manifolds, on $X$. But we may not assume $X$ to be a closed and orientable manifold. Thanks!

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    $\begingroup$ Do you have any additional hypotheses about $X$? $\endgroup$ – Eric Wofsey Jan 21 '16 at 2:30
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This is true if you assume $H_n(X)$ is finitely generated for all $n$ (all coefficients in this post will be $\mathbb{Z}$). In particular, this holds if $X$ has the homotopy type of a CW-complex with finitely many cells in each degree.

To prove this, we invoke the classification of finitely generated abelian groups, which says that $H_n(X)$ is a finite direct sum of cyclic groups. Since the universal coefficient theorem says that $H^n(X)\cong \operatorname{Hom}(H_n(X),\mathbb{Z})\oplus\operatorname{Ext}(H_{n-1}(X),\mathbb{Z})$ and Hom and Ext both preserve (finite) direct sums, to compute $H^n(X)$ we just have to compute $\operatorname{Hom}(A,\mathbb{Z})$ and $\operatorname{Ext}(A,\mathbb{Z})$ when $A$ is a cyclic group. In particular, we can use the following facts: $$\operatorname{Hom}(\mathbb{Z},\mathbb{Z})\cong \mathbb{Z}$$ $$\operatorname{Hom}(\mathbb{Z}/m,\mathbb{Z})\cong 0$$ $$\operatorname{Ext}(\mathbb{Z},\mathbb{Z})\cong 0$$ $$\operatorname{Ext}(\mathbb{Z}/m,\mathbb{Z})\cong \mathbb{Z}/m$$

It follows that $\operatorname{Hom}(H_n(X),\mathbb{Z})$ is a direct sum of copies of $\mathbb{Z}$, one for each direct summand of $\mathbb{Z}$ in $H_n(X)$, and $\operatorname{Ext}(H_{n-1}(X),\mathbb{Z})$ is a direct sum of finite cyclic groups which are isomorphic to the finite cyclic groups appearing as direct summands in $H_{n-1}(X)$. So, in brief, $H^n(X)$ is isomorphic to the direct sum of the free part of $H_n(X)$ and the torsion part of $H_{n-1}(X)$ (where "free part" means the direct sum of all the $\mathbb{Z}$ summands, and "torsion part" means the direct sum of the finite cyclic summands).

In particular, from the description above, we see that the torsion part of $H^n(X)$ is isomorphic to the torsion part of $H_{n-1}(X)$, so $H^n(X)$ has an element of order $p$ iff $H_{n-1}(X)$ has an element of order $p$. Considering the (co)homology in all degrees at once, we get that $H^*(X)$ has an element of order $p$ iff $H_n(X)$ has an element of order $p$.

Without the assumption that $H_n(X)$ is finitely generated, this need not be true. For instance, it is possible to construct a space $X$ such that $H_0(X)=\mathbb{Z}$, $H_1(X)=\mathbb{Q}/\mathbb{Z}$, and $H_n(X)=0$ for all $n>1$. Then the homology of $X$ has $p$-torsion for any $p$, but the cohomology can be computed to have no $p$-torsion: $H^0(X)=\mathbb{Z}$, $H^1(X)=0$, $H^2(X)=\hat{\mathbb{Z}}$, and $H^n(X)=0$ for $n>2$.

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  • $\begingroup$ Nice counterexample. I was trying to remember how $\text{Ext}^1(-, \mathbb{Z})$ behaved in general but couldn't immediately remember the relevant fact, namely that it acts as Pontryagin duality on torsion abelian groups. So we need to find a torsion abelian group whose Pontryagin dual is torsion-free and $\mathbb{Q}/\mathbb{Z}$ has this property. $\endgroup$ – Qiaochu Yuan Jan 21 '16 at 3:08
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These conditions are equivalent if $X$ is a levelwise finite CW complex (finitely many cells of each dimension), since this condition ensures that each homology group is finitely generated. You can prove this using the universal coefficient theorem as described here; it ensures that the torsion subgroup of $H_k(X, \mathbb{Z})$ is isomorphic to the torsion subgroup of $H^{k+1}(X, \mathbb{Z})$, so the homology and cohomology have the same torsion, just in degrees shifted by $1$.

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