Can the Ricci tensor determine a manifold up to homeomorphism? I don't know how to ask this question. Maybe there are many improper  statements. 

Let $(M, g)$ and $(N, h)$ be two Riemannian manifolds and $f:M \to N$ a  bijection such that $\operatorname{Ric}(x)$ is equal to $\operatorname{Ric}(f(x))$ up to a coordinate transformation. Are $M$ and $N$ homeomorphic?

In fact, I feel "up to a coordinate transformation" is not suitable, because, locally, I feel (not sure) that I can choose suitable coordinates making $\operatorname{Ric}(x) = \operatorname{Ric}(f(x))$. But I also don't know how to ask it. I just want to know whether the Ricci tensor can determine a manifold up to homeomorphism.

This is not true. For example, not all Ricci-flat manifolds are homeomorphic. More explicitly, for any bijection $f$ between $\mathbb{R}^n$ and $(S^1)^n$, $\operatorname{Ric}(x) = \operatorname{Ric}(f(x)) = 0$, but $\mathbb{R}^n$ and $(S^1)^n$ are not homeomorphic.
