Show that the set $\Bbb R^2 \setminus \{-e_1, e_1\}$ is homotopy equivalent with $S(-e_1, 1) \cup S(e_1, 1)$. 
Show that the set $\Bbb R^2 \setminus \{-e_1, e_1\}$ is homotopy equivalent with $S(-e_1, 1) \cup S(e_1, 1)$.

So this is asking to show that $\Bbb R^2$ with two points missing is homotopy equivalent with the "figure eight". To show this I need to find maps $f : \Bbb R^2 \setminus \{-e_1, e_1\} \to S(-e_1, 1) \cup S(e_1, 1) $ and $g : S(-e_1, 1) \cup S(e_1, 1) \to \Bbb R^2 \setminus \{-e_1, e_1\}$ that satisfy $g \circ f \simeq id_{\Bbb R^2 \setminus \{-e_1, e_1\}}$ and $f \circ g \simeq id_{ S(-e_1, 1) \cup S(e_1, 1)}$.
I'm very confused on how to start constructing such maps. It seems as if I would need to map the whole plane to $S(-e_1, 1) \cup S(e_1, 1)$ somehow. Can I use the inclusion map here to map all points of the plane to $S(-e_1, 1) \cup S(e_1, 1)$?
 A: You want your map g to be the inclusion map. For f you want to use the gluing lemma, to find fitting subsets of $\mathbb{R}^2$, which can be send to the figure eight (for example you can choose one of the subsets to be $B_1(-e_1)\backslash \{-e_1\}$, and find a map that gives a homotopy equivalence between $S(-e_1,1)\cong B_1(-e_1)\backslash S(-e_1,1)$
A: Assume that the circles are $\mathbb S^1+(0,1)$ and $\mathbb S^1+(0,-1)$. Put $c=(0,1)$ and consider
$$
    r_1\colon\{(x,y)\mid y\ge0\}\setminus\{c\}\to X
$$
defined as illustrated here

Do something similar for $y\le0$ and then define $r\colon\mathbb R^2\setminus\{c,-c\}\to X$.
It remains to be shown that $r$ is retraction by deformation. Consider
\begin{align*}
    H\colon\mathbb R^2\setminus\{c,-c\}\times[0,1]
        &\to\mathbb R^2\setminus\{c,-c\}\\
    ((x,y),\tau)&\mapsto(1-\tau)r(x,y) + \tau(x,y).
\end{align*}
To see that $H$ is well defined one needs to verify that equality in the equation
\begin{equation}
    \pm(0,1) = (1-\tau)r(x,y)+\tau(x,y)
\end{equation}
is never attained.
