Weakest condition for a locally Hausdorff space to be (globally) Hausdorff?

Question

Definition - "locally Hausdorff": every point $x$ of the topological space $X$ has a neighborhood $U$ which is Hausdorff in the subspace topology, equivalently is homeomorphic to a Hausdorff space.

Example: any locally Euclidean space is locally Hausdorff, since Euclidean spaces are Hausdorff.

Clearly being locally Hausdorff is not sufficient to be Hausdorff, as the line with two origins or any other non-Hausdorff manifold will show.

Question: Are there any conditions, either weaker than being globally Hausdorff or incomparable with being globally Hausdorff, which one could impose on a locally Hausdorff space to ensure that it is globally Hausdorff?

In particular, are there any local conditions which could be imposed?

A reference or pointers to references will suffice for an answer.

For example, the line with two origins only fails to be Hausdorff because of two neighborhood bases. Thus, based on this one example, it seems like it should be possible to impose a local condition on the neighborhood bases of a locally Hausdorff space to ensure global Hausdorffness.

In any case, imposing global Hausdorffness directly seems like it should be more than necessary somehow. Maybe it is not, and if it isn't, I would like to learn and understood how it is necessary.

Answer 1

Maybe the following could be usefull: If X is a topological group, then it is Hausdorff if it is $T_1$. So in particular, it is Hausdorff if it is locally Hausdorff. Since we can translate neighborhood bases, we only need to check local Hausdorffness at a single point. Thus a non-Hausdorff space which is locally Hausdorff at some point cannot carry a group structure.