## Homotopy type of $\mathbb{R}^3 \setminus \{ \mathrm{wedge \hspace{3pt} sum \hspace{3pt} of \hspace{3pt} 2 \hspace{3pt} circles} \}$

Is there something “nice” that the space $$\mathbb{R}^3 \setminus \{ \mathrm{wedge \hspace{3pt} sum \hspace{3pt} of \hspace{3pt} 2 \hspace{3pt} circles} \}$$ is homotopy equivalent to? I know that $$\mathbb{R}^3 \setminus \{\mathrm{circle}\} \simeq S^{2} \vee S^{1}$$, and that $$\mathbb{R}^3 \setminus \{ \mathrm{2 \hspace{3pt} linked \hspace{3pt} circles}\} \simeq S^{2} \vee T^{2}$$. However, I can’t seem to apply either of the “visual methods” used to prove these homotopy equivalences to my case.

My wild guess is that it might be homotopic to $$M_{2}$$, the orientable closed surface of genus $$2$$, but I’m not sure if this is true at all.