Retracting $\mathbb{R}^3$ minus the wedge sum of two circumferences to a double torus.

Question

I am trying to prove that there is a deformation retraction from $\mathbb{R}^3-\left(S^1\vee S^1\right)$ to a double torus.

I don't need a super rigurous proof, but I am not able to picture it.

Answer 1

Begin by collapsing $\mathbb{R}^3$ to $B$, where $B$ is a ball containing $S^1 \vee S^1$. Next, slightly engulf $S^1 \vee S^1$ to obtain an $\epsilon$-neighbourhood, call it $N$ (picture $\epsilon$ as being very small). So we are left with $B-N$. Note that $N$ is just a solid double torus. Imagine poking a straw into $B-N$ and sucking all the air out. Then the whole thing collapses onto the boundary of $N$, aka a double torus.

In summary: \begin{align*} \mathbb{R}^3-(S^1\vee S^1) \mapsto B-(S^1 \vee S^1) \mapsto B-N \mapsto \text{double torus}, \end{align*} where the first map is collapsing $\mathbb{R}^3$, the second map is engulfing $S^1 \vee S^1$, and the final map is sucking the air out of $B-N$.