general relativity - How to make sense of $mathcal{I}^-$ as a Cauchy surface rigorously?

In some references, like Hawking's derivation of black hole radiation, one considers that $\mathcal{I}^-$ is a Cauchy surface. One recent reference with such a claim is the paper "Soft Hair on Black Holes" (page 7 after equation 2.8), in verbis:

In the absence of stable massive particle or black holes, $\mathcal{I}^+$ ($\mathcal{I}^-$) is a Cauchy surface

Well by definition:

A Cauchy Surface for the spacetime $(M,g)$ is a surface $\Sigma\subset M$ such that every inextendible causal curve on the structure $(M,g)$ hits $\Sigma$ exactly once.

So part of the definition requires $\Sigma$ to be a subset of spacetime points. But strictly speaking in a rigorous framework $\mathcal{I}^-$ is not a place of the physical spacetime $(M,g)$. It is actually defined just on the conformal completion to one unphysical spacetime $(\hat{M},\hat{g})$ as the boundary $\mathcal{I}=\partial \hat{M}$.

So when one says that $\mathcal{I}^{-}$ is a Cauchy surface, how can we actually understand that statement from a rigorous standpoint?

Does it mean that when $(M,g)$ is globally hyperbolic so is any unphysical conformal completion $(\hat{M},\hat{g})$ and that any inextendible causal curve in $(M,g)$ when extended with respect to the $(\hat{M},\hat{g})$ structure (on which it is not inextendible) will hit $\mathcal{I}^-$ exactly once?

This seems like a problem because timelike geodesics doesn't hit $\mathcal{I}^-$. So what is going on here in the end?