Let $(M,g)$ be a time-oriented smooth Lorentzian manifold, with Lorentzian metric $g$. In the following thread:

physicist @ValterMoretti makes the following claim:

"As a matter of fact, a (connected) Lorentzian smooth metric $g$ over the time-oriented smooth manifold $M$ does define a topology, the same already present on $M$."

I have not been able to find a proof of the statement above, unless some relatively strong assumptions are made on $(M,g)$ (for example imposing that $(M,g)$ is strongly causal). Can the statement be proved in general without further extra-assumptions? I was under the impression that in fact one of the key points of departure between Riemannian and Lorentzian geometry is the failure for a Lorentzian metric to define in general a topology equivalent to the pre-existing manifold topology, in contrast to what happens in Riemannian geometry.

Thanks.