Are a two-petal rose and a Hopf link with joined circles topologically equivalent?

Question

I have a very limited knowledge on topology. From what I can visualize in my mind, a 2-petal rose (or a bouquet of 2 circles, or an 8 figure) in 3 dimensions, is equivalent or homeomorphic to a Hopf link where the two circles are joined. These two circles being joined means that for example a line segment links the part of circle A which is furthest away from the circle B with the point in circle B which is furthest away from circle A.

I say that I can visualize it in my mind because I think I can imagine the line segment being first moved into the points of the circles which are closest to one another, then deformed into a point, and then the circles' link can be reduced to this point and thus obtain a figure eight.

However, if I think in loops over one circle of the eight figure, I can make as many loops as I want wihout ever going through the other circle. In the joined Hopf link, if I draw loops over one of the circles I pass through the inside of the other circle once every loop. I'm not able to imagine how to draw loops in the joined Hopf link to avoid going through the other circle.

I don't know how this "going through the other loop" is defined topologically, but it seems that is should be something worth making the two figures topologically inequivalent. However, as explained in the first paragraph it seems to me they should be. If both figures are equivalent, how is this "going through the other loop" achieved in the eight figure?

Answer 1

There are several notions of “topological equivalence” of subspaces $A$ and $B$ of a space $X$, which we list from the strongest to the weakest.

• There exists a homeomorphism $h$ of the space $X$ such that $h(A)=B$. For instance, I guess, that if $X=\Bbb R^3$, $A$ consists of two linked circles and $B$ consists of two unlinked circles then there is no homeomorphism $h$ of $X$ such that $h(A)=B$.
• $A$ is homeomorphic to $B$. By the way, the figure 8 is not homeomorphic a Hopf link because 8 contains a point after the removal of which the figure becomes a disjoint union of open segments, whereas the Hopf link has no such points.
• $A$ is homotopy equivalent to $B$. According to “Algebraic Topology” Allen Hatcher

“the three graphs are all homotopy equivalent since they are deformation retracts of the same space”.