TITLE: Map from Integral Cohomology to Real Cohomology
QUESTION [6 upvotes]: 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. Thanks in advance!

REPLY [4 votes]: 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 $H^k(X;\mathbb{R})$.