Let $X$ be a topological Space and let $C^*(X)$ denote its singular cochain complex.
Since the cup-product of two cocyles is again a cocycle, we get a induced product structure on the complex of cocycles $Z^*(X)$.
By definition of this product we get the dga Morphism $Z^*(X) \to C^*(X)$ induced by the inclusion $Z^n(X) \subset C^n(X)$, which should be a quasi-isomorphism, by its definition.
Furthermore we can just define a map $Z^*(X) \to H^*(X)$ by mapping $z \in Z^*(X)$ to its cohomology class $[z]\in H^*(X)$. Since the cup-product for cohomology is induced by the cup-product for cochains, this map should also be dga morphism, which induces isomorphisms in homology (by viewing $H^*(X)$ as a dga with zero differential).
Therefore every topological space should be formal, which of course isn't true.
Thus one of those morphisms cant be a dga morphism, but I dont see any problem with those morphisms.
Best Answer
They are dga morphisms, but neither $Z^*(X)\to C^*(X)$ nor $Z^*(X)\to H^*(X)$ is a quasi-isomorphism.
$Z^*(X)$ is a dga with no differentials, so it's its own (co)homology ! (everyone in $\mathrm{im}(d)$ must come from someone whose $d$ is nonzero, that is, someone not in $Z$ !)