I'm trying to solve the following exercise:
I think I've understood the "union of two circles with a point in common" case but I'm lost on the other one. I tried to come up with examples of a retraction $r: \mathbb{R}^2 \setminus \{p, q\} \to \mathbb{S}^1 \cup \text{diameter} $ but I couldn't come up with any. The usual retraction $\mathbb{R}^2 \setminus \{p\}$ onto a circle with center $p$ doesn't work either, because in this case we have to attach a diameter. Can anyone help me? I'd appreciate it!
Hint: such a homotopy may be not so fun to make explicit. But here's an idea: if you collapse the diameter to a point, continuously, you get the two circles glued in a point as before.
What may work is picking a "center" $x,y$ in each half circle to remove, and sending each point $z$ to the intersection of the line $\vec{xz}$ (or $\vec{yz}$, depending on which half of the circle lies $z$) with $S^1$ Then for the outside of the circle we do "the same", fix $o$ to be the center of the circle and send some point $z$ outside the circle to $S^1 \cap \vec{zo}$.
That should define a continuous retraction $r$; we're in the plane so for concrete $o,x,z$ we can cook up a concrete formula and all, but I don't know if the homotopy $ir \simeq 1$ is easy to describe. My intuition is that since our retraction was defined by assigning each point $p$ a curve $\gamma_p$ from $p$ to $r(p)$, the homotopy should look like $H(t,p) = \gamma_p(t)$. Again, this has to be made precise.