MHB How are these spaces homotopy equivalent?

[Chris L T521]

Here's this week's problem.

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Problem: Show that the following spaces are homotopy equivalent:\[\mathbb{R}^3\backslash\{\text{two lines through the origin}\} \simeq S^2\backslash\{\text{four points}\}\ \simeq \mathbb{R}^2\backslash\{\text{three points}\}\simeq S^1\vee S^1\vee S^1.\]
Note that the last space ($S^1\vee S^1\vee S^1$) is a one point union of three circles.
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[Chris L T521]

No one answered this week's question. Here's my solution:

Spoiler

Proof: We begin with showing the first set of homotopy equivalences. Consider two lines that pass through the origin in $\mathbb{R}^3$. WLOG, let's take these lines to be two of the three standard coordinate axes. If we contract the entire space to a sphere, we observe that when we remove these two lines, we end up with the sphere being punctured in four different places; thus, we're left with \[\mathbb{R}^3\backslash\{\text{two lines through the origin}\}\simeq S^2\backslash\{\text{four points}\}.\]
Now, consider the sphere with four points removed. We can continuously deform this space by expanding one of the holes to the equator of the sphere, leaving us with a half sphere with three holes. Flattening this onto the plane gives us an open disc with three points. We can continuously expand the boundary of this disk out to infinity, thus leaving us with the homotopy equivalence
\[S^2\backslash\{\text{four points}\}\simeq \mathbb{R}^2\backslash\{\text{three points}\}.\]
(Another way to see this is by having one of the holes at the north pole, and then applying stereographic projection, leaving us with $\mathbb{R}^2$ minus three points.)

Now, consider $\mathbb{R}^2$ with three points removed. This is homotopic to an open disk with three points removed. If we expand these three holes to fit inside this open disk, we end up with an open disk split into three sections. Each of these sections intersect at one point; since each section is homotopic to $S^1$, we have
\[\mathbb{R}^2\backslash\{\text{three points}\}\simeq S^1\vee S^1\vee S^1.\]
Thus, we have the homotopic equivalence
\[\mathbb{R}^3\backslash\{\text{two lines through the origin}\} \simeq S^2\backslash\{\text{four points}\}\ \simeq \mathbb{R}^2\backslash\{\text{three points}\}\simeq S^1\vee S^1\vee S^1.\]
Q.E.D.