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I am reviewing for my topology final and came up with this example. I want to compute the homology groups of $X = \mathbb{R}^3 - \{C_1,C_2\}$ where $C_1$ and $C_2$ are disjoint copies of $S^1$, so basically the complement of two disjoint unknots.

I usually try to deformation retract $X$ to something easier to work with, but this time I'm having quite a hard time. What is the best way to do this? I would like to deformation retract $X$ so I can find its homology groups.

This example comes from one we did in class, namely $Y = \mathbb{R}^3 - \{C_1,C_2\}$ where $C_1$ and $C_2$ are copies of $S^1$ that are linked.

Hubble
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Using a deformation retract argument: $Y = \mathbb{R}^3 - \{C_1,C_2\}$ where $C_1$ and $C_2$ are unlinked. You can see that $Y$ deformation retract on a the wedge sum of two copies of $X=B^{3}- S^{1}$ (where $B^3$ is a $3-$ball) so you can just look at $X$.

$$X=B^{3}- S^{1}\simeq S^{3}-(S^{1}\cup\{p\})\simeq\mathbb{R}^3-(\mathbb{R}\cup\{p\})$$

This last space deformation retract on $S^{1}\vee S^{2}$.

Hence your space $Y$ is homotopy equivalent to $S^{1}\vee S^{1}\vee S^{2}\vee S^{2}$.

Dario
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Suppose that the circles are in the $xy$-plane, well to the left and right of the $x = 0$ plane. Then you can let $U = X \cap \{(x, y, z) | x < 1\}$ and $V = X \cap \{ (x, y, z) | x > -1\}$, and apply Mayer-Vietoris. Since the reduced (co)homology of $U \cap V$ is trivial, things get pretty easy in the MV-sequence.

(I know that's not a deformation-retract argument, but it's not a bad technique to know...)

John Hughes
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  • Seems like the way to go! I don't think I can get something interesting with a deformation retraction, but an MV-sequence gives it. Thanks! – Hubble Dec 04 '14 at 04:50