# Map from Integral Cohomology to Real Cohomology

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.

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 $$H^k(X;\mathbb{R})$$.

• It's not correct that cochains with coefficients in $\mathbb{R}$ are $C^\bullet(X)\otimes\mathbb{R}$. This works for chains but not cochains (since cochains are a product of copies of $\mathbb{Z}$, not a direct sum). Aug 9, 2021 at 5:50
• Similarly, $H^k(X;\mathbb{Z})\otimes\mathbb{R} \cong H^k(X;\mathbb{R})$ is not true in general (only when the cohomology is finitely generated). Aug 9, 2021 at 5:52
• @EricWofsey Oh, you're quite right, thanks. Aug 9, 2021 at 5:53

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 $$H^k(X;\mathbb{R})$$.