Homotopy type of R^3\(circles)

$\mathbf{1}$. The spaces $X=\mathbb{R}^3\setminus S^1$ and $S^{1}\vee S^{2}$ are homotopy equivalent.

First we rewrite $X\cong S^3\setminus (\{*\}\cup S^1)$, as $S^3$ is the one-point compactification of $\mathbb{R}^3$. Now observe $S^3\setminus S^1$ is homeomorphic to the open solid torus, $S^1\times e^2$, where $e^2$ is the open unit $2$-disc. An explicit homemorphism is as follows : $$(x,y,z,t)\mapsto (x\sqrt{1-z^2-t^2},y\sqrt{1-z^2-t^2},z,t).$$

 Thus $\mathbb{R}^3\setminus S^1\cong S^1\times e^2\setminus \{*\}$. Here $S^1\times e^2$ is open $3$-manifold. Removing a point from open $3$-manifold $M$ is wedging $M$ with $S^2$ $-$ we may think of enclosing the point with $S^2$ and retracting the enclosed region to the sphere (or blowing up the point to a sphere); then we may push the sphere away from the manifold, to the boundary, forming the wedge product. Thus $\mathbb{R}^3\setminus S^1\cong S^1\times e^2\vee S^2\sim S^1\vee S^2$. 

 

$\mathbf{2}$. The space $X=\mathbb{R}^3\setminus M$, where $M$ is the union of two disjoint unlinked circles, is homotopy equivalent to $S^{1}\vee S^1\vee S^{2}\vee S^2$.

As before, $X\cong S^1\times e^2\setminus(\{*\}\cup S^1)$. Observe that the remaining circle to be removed is contractible in $S^1\times e^2$, i.e. it is not encircling the hole of the torus. Thus we may take a $2$-sphere which bounds a region containing the circle to be removed. This regions retracts to $S^2\vee S^1$, as in $\mathbf{1}$. Pushing it to the boundary of $S^1\times e^2$ we get a wedge with $S^2\vee S^1.$ Removing the point $*$ induces, as in $\mathbf{1}$, one more wedge, with $S^2$. Hence $X\sim (S^1\times e^2)\vee S^2\vee S^2\vee S^1$, which finishes the derivations as $S^1\times e^2\sim S^1$

 

$\mathbf{3}$. The space $X=\mathbb{R}^3\setminus M$, where $M$ is the union of two disjoint linked circles, is homotopy equivalent to $S^{2}\vee (S^1\times S^1)$.

As in $\mathbf{2}$, $X\cong S^1\times e^2\setminus(\{*\}\cup S^1)$. Observe that the remaining circle to be removed is not contractible in $S^1\times e^2$, i.e. it encircles the hole of the torus. Removing the point $*$ induces, as in $\mathbf{1}$, a wedge with $S^2$.  Hence $X\sim ((S^1\times e^2)\setminus S^1)\vee S^2$. But $(S^1\times e^2)\setminus S^1$ is a solid torus with core circle removed, which clearly deformation retracts to a torus $-$ $S^1\times S^1$.