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.