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Background:

I recently discovered that the complement to the circle and vertical axis shown below is homotopy equivalent to a torus

torus torus2

Also complement to three infinite straight non-intersecting lines in $\mathbb{R^3}$ as illustrated below is homotopy equivalent to a sphere with $3$ holes drilled through (realised by retracting $\mathbb{R^3}$ onto the open ball)

lines sphere


My question:

I was curious to know if we could represent two other common topological spaces: the Klein bottle and projective plane with simple diagrams as above which they are homotopic too.

Could someone provide me with examples of these? (I hope this question is OK and not too broad)

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1 Answer 1

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I interpret your question as follows: "Is there a (nicely) embedded graph in $S^3$ whose complement is homotopy equivalent to the projective plane or Klein bottle?" To see why both of your examples fall under this, think of $S^3$ as $\Bbb R^3$ with a point at infinity. Then your first example is the complement of two circles in $S^3$ (the picture you get is called the Hopf link, and your second example the complement of a graph with one vertex and three edges.

Note, by the way, that because these spaces do not embed in $\Bbb R^3$, then certainly your open set cannot deformation retract onto them. This gives us the hint that no, what you want is not possible.

The theorem we want is Alexander duality (though we'll dualize in the opposite way that Wikipedia does) - this says that for your graph $X$, $H^1(X) = H_1(S^3 \setminus X)$. Graphs can never have torsion in either their $H_1$ or their $H^1$, and hence $H_1(S^3 \setminus X)$ is torsion-free. But $H_1$ of a non-orientable surface always has a 2-torsion element! So this is impossible. It's much harder to prove, but still true, that an arbitrary open subset of $S^3$ cannot have torsion in its first homology group, so even a more general formulation of your question is still not possible.

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  • $\begingroup$ I was curious to know if the Klein bottle and projective plane were homotopy to some union of common spaces, such as lines and circles as in the 2 examples given... do you think this is not possible? $\endgroup$
    – thinker
    May 24, 2016 at 23:19
  • $\begingroup$ @thinker This answer contains a proof that this is not possible. $\endgroup$
    – user98602
    May 24, 2016 at 23:19
  • $\begingroup$ right ok, its just your answer seems too advanced for me $\endgroup$
    – thinker
    May 24, 2016 at 23:20
  • $\begingroup$ @thinker It tends to be hard to prove that something isn't possible! You're going to need some machinery to do this. The tools of algebraic topology (including homology, ever-present in the above answer) are what we need - there's not going to be a proof without it. $\endgroup$
    – user98602
    May 24, 2016 at 23:21
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    $\begingroup$ Nicely embedded is there because I doubt the OP has in mind things like the complement of this fella. I already clarified what we're embedding: a graph, something made with finitely many vertices and finitely many edges. I'm just making sure we haven't embedded it in an absurd way like the above (though this is not necessary for the argument). The cylinder is homotopy equivalent to the circle, so just take the complement of the z-axis. $\endgroup$
    – user98602
    May 24, 2016 at 23:26

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