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.