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I know the answer for the question if the embedding in the question is from $I=[0,1]$ into $\mathbb{S}^2$ or from $\mathbb{S}^1$ into the sphere. It is direct from the Proposition 2B.1 from Chapter 2B of Hatcher Algebraic Topology.

However, I want to know the answer if $I=(0,1)$ is open. In more detail, if we have a simple non-closed curve in the sphere, what is the homology group of the complement of that curve.

EDIT: A friend of mine has showed me a proof that such a curve would divide the sphere into one or two connected components. I am not very familiar with algebraic topology but I suppose that means the group is either $0$ or $\mathbb{Z}$.

Jose Pérez Cano
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  • I would suspect the complement has the homotopy type of either ${}$ or ${} \sqcup {*}$, but I am not sure. – Connor Malin Nov 10 '21 at 16:39
  • @ConnorMalin there are many different possibilities. For example the "$\infty$" figure can be created from an injective curve $(0,1)\to \mathbb{S}^2$ which divides $\mathbb{S}^2$ into $3$ connected components. In fact I'm pretty sure that we can find a curve that divides $\mathbb{S}^2$ into arbitrarily (finitely) many components. Although I don't know if there is an example of such decomposition where at least one component is not contractible. – freakish Nov 11 '21 at 00:06
  • @freakish the infinite curve won't work. If the curve is from $(0,1)$ which is open, then you only have one connected component. Although you have 3 path-connected components. That is because the infinite shaped curve isn't closed. – Jose Pérez Cano Nov 11 '21 at 16:45
  • The image of the $\infty$ curve is closed, it cuts the sphere into three connected components. The "missing endpoints" are included by the points in the middle portion of the curve. – Ethan Dlugie Nov 11 '21 at 16:58
  • The complement to the image is, in general, not open, so path-compinents are not the same as components for the complement. Or maybe you assume that the image of your map is a closed subset of the sphere? Also, you should clarify if you want a map to be a topological embedding or a continuous injection. People routinely confuse the two notions. – Moishe Kohan Nov 11 '21 at 19:27
  • See also my answer here. – Moishe Kohan Nov 11 '21 at 20:03

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