Map from Integral Cohomology to Real Cohomology

Question

Let $X$ be a topological space. For each $k \in \mathbb{Z}$ we can consider the following map.

$$ j_k : \mathrm{H}^k(X;\mathbb{Z}) \to \mathrm{H}^k(X;\mathbb{R}) \ ; \ [f]_{\mathbb{Z}} \mapsto [f]_{\mathbb{R}}. $$

(To reduce confusion, let us denote things in $\mathrm{H}^k(X;\mathbb{Z})$ by subscript $\mathbb{Z}$, and things in $\mathrm{H}^k(X;\mathbb{R})$ by subscript $\mathbb{R}$.) It is not hard to show that this map is a well-defined homomorphism. I want to know when this map is injective. Of course, if $\mathrm{H}^k(X;\mathbb{Z})$ is not torsion-free, the map is not injective. Then is it true that the map is injective if $\mathrm{H}^k(X;\mathbb{Z})$ is torsion-free?

I was first trying to prove for the simplest case that $X$ is a compact connected orientable smooth manifold and $k = \dim X$ so that $\mathrm{H}^k(X;\mathbb{Z}) \cong \mathbb{Z}$ and $\mathrm{H}^k(X;\mathbb{R}) \cong \mathbb{R}$. In this case, if we say $[f]_{\mathbb{Z}} \in \mathrm{H}^k(X;\mathbb{Z})$ is a generator, then the map $j_k$ is completely determined by $j_k([f]_{\mathbb{Z}}) = [f]_{\mathbb{R}} \in \mathrm{H}^k(X;\mathbb{R})$. Therefore, the map $j_k$ must be either injective or zero, but I do not know whether the map must be nonzero.

I would very appreciate even for partial answers.

Answer 1

If $H^k(X;\mathbb{Z})$ is finitely generated (like in the case for a compact smooth manifold), the universal coefficient theorem for cohomology (Spanier chapter 5.5 theorem 10) gives short exact sequences: $$0 \to H^k(X;\mathbb{Z})\otimes\mathbb{R} \to H^k(X;\mathbb{R}) \to \operatorname{Tor}_1^{\mathbb{Z}}(H^{k+1}(X;\mathbb{Z}),\mathbb{R}) \to 0$$ Since $\mathbb{R}$ is a flat $\mathbb{Z}$-module, the tor term vanishes, so the short exact sequences give rise to isomorphisms: $$H^k(X;\mathbb{Z})\otimes\mathbb{R} \cong H^k(X;\mathbb{R})$$ In fact, this isomorphism is defined by the rule $[f]_{\mathbb{Z}}\otimes c\mapsto c[f]_{\mathbb{R}}$.

Given an abelian group $A$, the homomorphism $A\to A\otimes\mathbb{R}$ defined by $a\mapsto a\otimes 1$ has kernel the torsion subgroup $T$ of $A$, so the image is the free abelian group $A/T$. By choosing a generating set for $A/T$, one can check that $A\otimes\mathbb{R}$ is isomorphic to $(A/T)\otimes\mathbb{R}$, a vector space over $\mathbb{R}$ with a basis is given by that generating set.

To answer your question, the kernel of $j_k$ is the torsion subgroup of $H^k(X;\mathbb{Z})$, and the image is the free part, the span of which over $\mathbb{R}$ is all of $H^k(X;\mathbb{R})$.

Answer 2

The universal coefficients theorem gives an isomorphism (natural in the coefficient group $A$) $$H^k(X;A)\cong \operatorname{Hom}(H_k(X),A)\oplus \operatorname{Ext}(H_{k-1}(X),A)$$ (here homology is always with coefficients in $\mathbb{Z}$). So the natural map $H^k(X;\mathbb{Z})\to H^k(X;\mathbb{R})$ is injective iff the maps $\operatorname{Hom}(H_k(X),\mathbb{Z})\to \operatorname{Hom}(H_k(X),\mathbb{R})$ and $\operatorname{Ext}(H_{k-1}(X),\mathbb{Z})\to \operatorname{Ext}(H_{k-1}(X),\mathbb{R})$. The first map is always injective since it is just composition with the inclusion $\mathbb{Z}\to\mathbb{R}$. The second map is always $0$ since $\operatorname{Ext}(H_{k-1}(X),\mathbb{R})=0$ (since $\mathbb{R}$ is an injective abelian group).

So, $H^k(X;\mathbb{Z})\to H^k(X;\mathbb{R})$ is injective iff $\operatorname{Ext}(H_{k-1}(X),\mathbb{Z})$ is trivial. When $H_{k-1}(X)$ is finitely generated, this Ext group is always torsion, and so it follows that $H^k(X;\mathbb{Z})\to H^k(X;\mathbb{R})$ is injective if $H^k(X;\mathbb{Z})$ is torsion-free. This applies, for instance, to compact manifolds, or finite CW-complexes. In general, though, $\operatorname{Ext}(H_{k-1}(X),\mathbb{Z})$ can be nontrivial but torsion-free. For instance, if $H_{k-1}(X)\cong\mathbb{Q}$ then $\operatorname{Ext}(\mathbb{Q},\mathbb{Z})$ is nontrivial (in fact, uncountable; see here for instance) but is a $\mathbb{Q}$-vector space. So if you take a space $X$ with $H_{k-1}(X)\cong\mathbb{Q}$, then $H^k(X;\mathbb{Z})$ will be torsion-free but will have a huge direct summand coming from $H_{k-1}(X)$ that is killed when mapped to $H^k(X;\mathbb{R})$.