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I am thinking whether there is a simple space equivalent to $R^3$ with a figure 8 removed from xy-plane. And what about wedge sum of n circles is removed?

For only one circle, the complement space actually is homotopic equivalent to a circle wedge $S^2$, there is a picture for this in Hatcher. So what about other cases?

user330928
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    The other case are similar. If you remove a wedge of $n$ circles, the complement is homotopy equivalent to a wedge of $n$ circles and one $S^2$. – Cheerful Parsnip Aug 18 '16 at 02:36

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Use the Wirtinger representation. Construct a $2$-complex by taking a bouquet of circles, one for each generator, and then gluing in a disc according to each relation. This can be nicely embedded in $\mathbb{R}^3$.

It would be what you got if you took a long balloon in the shape of a figure 8, inflated it, and then identified portions of the boundary that were pressed against each other.

Neal
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