Determining if the singular cohomology "change-of-coefficients" map for $\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}$ is surjective Let $H^k(-;A)$ denote the $k$th singular cohomology with coefficients in a (discrete / non-topological) Abelian group $A$.
I am trying to determine whether the change-of-basis map
$H^k(X;\mathbb{R})\rightarrow H^k(X;\mathbb{R}/\mathbb{Z})$,
induced by the canonical projection homomorphism $\mathbb{R}\twoheadrightarrow \mathbb{R}/\mathbb{Z}$,
is surjective. (In general, or at least for $X$ a smooth manifold.)
Would anyone know how to approach this problem? Any help is greatly appreciated.

What I have so far:
Given an $\mathbb{R}/\mathbb{Z}$-valued $k$-cocycle $\alpha \in\mathrm{Hom}_\mathbb{Z}( C_k(X) , \mathbb{R}/\mathbb{Z})$,
using that $C_k$ is free and hence projective,
we get an $\mathbb{R}$-valued cochain $\tilde{\alpha} \in \mathrm{Hom}_\mathbb{Z}(C_k(X),\mathbb{R})$ which is projected to $\alpha$ after post-composing by $\mathbb{R}\twoheadrightarrow \mathbb{R}/\mathbb{Z}$.
However, a priori this $\tilde{\alpha}$ is not a cocycle (i.e. vanishing on boundaries), instead $\tilde{\alpha}$ sends boundaries into $\mathbb{Z}$ (the kernel of $\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}$). Is it possible to show we can choose the lift $\tilde{\alpha}$ such that it is a cocycle and not just a cochain?
 A: If you want $H^k(X, \mathbb{R})$ to reproduce what people normally mean by "singular cohomology with coefficients in $\mathbb{R}$" then you need to give $\mathbb{R}$ the discrete topology; with the Euclidean topology $\mathbb{R}$ is contractible so cohomology with coefficients in $\mathbb{R}$ is trivial.
In general, if $0 \to A \to B \to B/A \to 0$ is a short exact sequence of abelian groups, then we get a coefficient long exact sequence in cohomology
$$\dots \to H^k(X, A) \to H^k(X, B) \to H^k(X, B/A) \xrightarrow{\beta} H^{k+1}(X, A) \to \dots $$
where the connecting homomorphism $\beta$ is a Bockstein homomorphism. Applied to the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{R}/\mathbb{Z} \to 0$ (of discrete groups, so we also give $\mathbb{R}/\mathbb{Z}$ the discrete topology) this produces a long exact sequence
$$\dots \to H^k(X, \mathbb{Z}) \to H^k(X, \mathbb{R}) \to H^k(X, \mathbb{R}/\mathbb{Z}) \xrightarrow{\beta} H^{k+1}(X, \mathbb{Z}) \to \dots$$
showing that the map $H^k(X, \mathbb{R}) \to H^k(X, \mathbb{R}/\mathbb{Z})$ is neither injective nor surjective in general; it's not injective in general since by exactness its kernel is the image of $H^k(X, \mathbb{Z}) \to H^k(X, \mathbb{R})$ which is generally nonzero (if everything in sight is finitely generated it's exactly the torsion-free quotient of $H^k(X)$) and it's not surjective in general since by exactness its image is the kernel of the Bockstein homomorphism $\beta$, which is generally nonzero (since by exactness again the image of $\beta$ is the kernel of $H^{k+1}(X, \mathbb{Z}) \to H^{k+1}(X, \mathbb{R})$ which again if everything is finitely generated is exactly the torsion subgroup of $H^{k+1}(X)$).
If on the other hand you want $\mathbb{R}$ and $\mathbb{R}/\mathbb{Z}$ to have the Euclidean topology then we still get a long exact sequence as above except that the terms $H^k(X, \mathbb{R})$ vanish as above, and we conclude that in this case the Bockstein just produces an isomorphism $H^k(X, \mathbb{R}/\mathbb{Z}) \cong H^{k+1}(X, \mathbb{Z})$, corresponding to the delooping isomorphism $B^k \mathbb{R}/\mathbb{Z} \cong B^{k+1} \mathbb{Z}$ of classifying spaces.
