TITLE: Embedding linklessly embeddable graphs without Borromean rings
QUESTION [5 upvotes]: A linklessly embeddable graph is a graph which can be embedded into $\Bbb R^3$ so that no two of its cycles are linked. For example, the Petersen graph is not such a graph.
Now, I can think of another type of "linkedness" that is not already addressed by this notion, namely, whether an embedding contain a Borromean rings configuration.
By that I mean three cycles of the graph, no two of which are linked, but all three can still not be "entangled into three separate cycles" (see the image from Wikipedia).
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Question: if a graph is linklessly embeddable, does it also have a linkless embedding without a Borromean rings configuration?
Or the other way around, are there linklessly embeddable graphs that cannot be linklessly embedded without Borromean rings?


Some thought
Maybe one can show that whenever a graph cannot be embedded without Borromean rings, then the following two (black) paths must be present. This would then already imply the presence of a link:

This does not address the possibility that a graph might be linklessly embeddable, and embeddable without Borromean rings, but not both at the same time.

REPLY [2 votes]: Isn't it that linklessly and flatly embeddable are the same family, and that a flat embedding can not contain a Borromean ring?
upd - clarification:
From the wiki article on linkless embeddings: "A flat embedding is an embedding with the property that every cycle is the boundary of a topological disk whose interior is disjoint from the graph." As far as I remember (and the wikipedia seems to confirm), every linklessly embeddable graph has a flat embedding (every flat embedding is trivially linkless). Finally, that a flat embedding can't contain a Borromean ring seems trivial to me from the above definition of a flat embedding (still can be wrong)