Consider a graph $G$ with at least two unavoidable crossings, say, the disjoint union of two copies of $K\_5$. Can such a graph always be drawn so that there is only one singular point (where all crossings happen)? I guess there is an easy proof that this is not possible.



Can all crossings in a graph be moved to one point?
**No**, this is not always possible.


**Lemma.** Let $G$ be an $n$-vertex graph with at least $3n-2$ edges. Then $G$ cannot be drawn in the plane so that all crossings occur at the same point.


*Proof.* We make the standard assumption that every pair of edges which intersect in a drawing are not 'tangent' at the point of intersection. Suppose $D$ is a drawing of $G$ where all edges cross at the same point. Let $G'$ be the graph obtained from $D$ by introducing a new vertex at the crossing point and then suppressing all parallel edges. We claim that $|E(G')| \geq |E(G)|$. Let $H$ be the subgraph of $G$ induced by the edges which pass through the crossing point. Since no two crossing edges are tangent, $H$ contains at most one cycle (which must be a triangle). Therefore, $H$ has average degree at most $2$. It follows that $|E(G')| \geq |E(G)|$, as claimed. Thus, $G'$ is a planar graph with $n+1$ vertices and at least $3n-2$ edges. But this contradicts the fact that every $n$-vertex planar graph has at most $3n-6$ edges.


In particular, the above lemma implies that $K\_7$ cannot be drawn so that all edges cross at the same point.


**Acknowledgement.** Many thanks to bof for help in making this answer converge to its present form (see the comments below).


