What singular link is this? A knot is a smooth map $f:\Bbb S^1\to\Bbb R^3$. A link is a collection of knots which do not intersect, but are linked together. A singular knot is a smooth map $f:\Bbb S^1 \to \Bbb R^3$ whose image has singularities.
First take $N$ longitudinal geodesics of one sphere $\Bbb S^2_1$ and map them to $\Bbb R^3$. Next take $N$ longitudinal geodesics of another sphere $\Bbb S^2_2$ and map them to $\Bbb R^3$. However $\Bbb S^2_2$ is rotated by $\pi/2$ radians. 
These structures are linked by alternately weaving over and under each other. There are singularities at the four poles where there are $N$ crossings. 
Is this link trivial or non-trivial? Is it prime or not prime? What questions could one ask about the complement of this structure?
Thanks.
 A: This is what I'm getting from your description:

This is a diagram from a stereographic projection near one of the four singular points.  The relationship to the number of geodesics is that $N=n+1$.  There is a bit of an ambiguity because there are two choices for each mesh --- is it "over-under" or "under-over"? --- but it doesn't seem to matter much for the questions you ask.
My interpretation of equivalence is that two of these things are equivalent if there is an isotopy carrying one to the other while preserving the singularities along the way.  This is what treating this as a four-vertex spatial graph would give us.
There is a question of what a trivial spatial graph ought to be.  Let's say it's whether there is a diagram for it without crossings.  If $n\geq 2$, then if you take the subgraph from first two rows and first two columns of one of the two meshes, you get the two-component link L4a1.  Thus, the spatial graph is nontrivial.  The $n=0$ case is trivial, and for the $n=1$ case it depends on what the two crossings are (the earlier ambiguity).   One case is trivial, but in the other the two loops composed only of mesh edges form a Hopf link.
What should prime mean?  For spatial graph theory, it means there is a sphere in $\mathbb{R}^3$ that intersects the graph in exactly two non-vertex points, where each side of the sphere is a non-trivial spatial graph in some sense. From the perspective of the graph itself, the points when removed must disconnect the graph into two pieces.  When $n\geq 1$, the graph is $3$-edge-connected, so the two cut points must lie on the same edge.  But any portion of any edge is not knotted (that's not rigourous; I have a longer proof involving looking at the intersection of a sphere with a checkerboard surface of the diagram), so the spatial graph is prime.  For $n=0$, it is just an unknot.
The complement of (a regular neighborhood of) this graph in $\mathbb{R}^3$ for $n=1$ is actually homeomorphic to the complement of an unknotted bouquet of $5$ circles, which can be identified as a genus-$5$ handlebody minus a ball.  The thing about graph complements is that you are allowed to slide the endpoint of an edge along another edge that it is incident to, so everything becomes the complement of a bouquet, though possibly knotted.  For $n\geq 2$, you can slide outermost rows and columns of the meshes one at a time to isolate circles of a bouquet.  Each mesh contributes $2n$ circles, and the remainder of the graph contributes a circle, and one can see that in general the complement is the complement of a bouquet of $4n+1$ unknotted circles.  Thus, the complement is a genus-$(4n+1)$ handlebody minus a ball.  (If the graphs were in $S^3$ rather than $\mathbb{R}^3$, there would not be the "minus a ball" part.)
