Lifting of a "homological character" to a differential form?

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For a topological Abelian group (assumed to have discrete topology), let $H_k(-;A)$ and $H^k(-;A)$ denote the usual ordinary singular homology and cohomology with coefficients in $A$.

For $X$ a topological space, we refer to elements of $\mathrm{Hom}_{\mathbb{Z}}(H_k(X;\mathbb{Z}),\mathbb{R}/\mathbb{Z})$ as "degree-$k$ homological characters".

(Note: by UCT, since $\mathbb{R}/\mathbb{Z}$ is an injective module over $\mathbb{R}$, it follows that deg-$k$ homological characters are naturally identified with $\mathbb{R}/\mathbb{Z}$-coefficient deg-$k$ cohomology classes.)


Now in what follows suppose $X$ is a smooth manifold.

Denote by $C_k(X)$ the usual $\mathbb{Z}$-coefficient singular chain complex of $X$; let $C^{\text{sm}}_k(X)$ be the subcomplex consisting of the smooth chains; and let $Z^{\text{sm}}_k(X) = C^{\text{sm}}_k(X) \cap Z_k(X)$ be the smooth cycles.

Let $\Omega^k(X)$ denote the space of differential $k$-forms on $M$.

I would like to show (or disprove) the following:

If $\alpha \in \mathrm{Hom}_{\mathbb{Z}}(H_k(X;\mathbb{Z}),\mathbb{R}/\mathbb{Z})$ is a degree-$k$ homological character on $X$, then there exists $\tilde{\alpha} \in \Omega^k(X)$, such that for all $c \in Z^{\text{sm}}_k(X)$ we have $$ \alpha([c]) = \left( \int_c \tilde{\alpha} \;\;\text{ mod }\mathbb{Z} \right) \,. $$


So far, I have the following:

Let us assume $H_{k}(X;\mathbb{Z})$ is a projective module over $\mathbb{Z}$. (In particular torsion-free.)

Then from $\alpha \in \mathrm{Hom}_{\mathbb{Z}}(H_k(X;\mathbb{Z}),\mathbb{R}/\mathbb{Z})$ we get a lift $\beta \in \mathrm{Hom}_{\mathbb{Z}}(H_k(X;\mathbb{Z}),\mathbb{R})$, unique up to addition by a $\mathbb{Z}$-valued homomorphism.

And, since $\mathbb{R}$ is injective over $\mathbb{Z}$ and by UCT (the Ext term in UCT vanishes) we get a unique $\gamma \in H^k(X;\mathbb{R})$ corresponding to $\beta$; by de Rham theory we can identify $\gamma$ with a de Rham class $\tilde{\alpha} \in H^k_{\mathrm{dR}}(X)$.

I was wondering, at this point how would we show directly the above property holds?