Infinite graph is planar iff it can be embedded in sphere

Question

My question is about the following statement about planar graphs:

A graph is planar (i.e. can be embedded in the plane) if and only if it can be embedded in the sphere $S^2$.

By an embedding we mean the following:

To every vertex $v \in V$ we associate a unique point in $\mathbb{R}^2$(or $S^2$). To every edge $e \in G$ we associate a unique simple arc, which is a homeomorphic image of $[0,1]$, connecting the points associated to its end vertices such that no two arcs intersect other than in a common vertex point.

The "only if" part of the statement above follows directly by using stereographic projection, which gives an embedding $\mathbb{R}^2 \hookrightarrow S^2$. For the "if" part, in every proof I find one simply says that in an embedding of a graph in $S^2$, one can always avoid a point and can therefore use stereographic projection again to get the embedding in $\mathbb{R}^2$.

I can see this fact being true for finite graphs, as the embedded graph is just the bijective continuous image of finitely many intervals glued together in some way (which is a compact space), so if the image was the whole $S^2$, we would get a homeomorphism between $S^2$ and something which isn't $S^2$.

But what about infinite graphs? As far as I know, a graph is just defined to be a tuple $(V,E)$, where $V$ is a set (of vertices) and $E$ is a set consisting of some 2-element-subsets of $V$. So $V$ can be basically anything, so I could for example just take $V = S^2$ (as a set), or any other (uncountable) infinite set. In this case, how can I ensure that in a drawing of the graph on $S^2$ I can still avoid one point? Or is this statement not even true for infinite graphs?

Answer 1

Regarding the comments on 'topological embedding' vs immersed realization: A 'topological graph' in all standard definitions, even for infinite graphs, defines a one-dimensional space (in the sense of covering dim, ind or Ind). Thus, as a one-dimensional subset of a surface (sphere) it does not contain any open set, by the standard result in dimension theory that the open 2-ball has dimension two and the result that dimension is non-increasing for subsets of separable metric spaces. Thus it misses a point of the sphere, which we may assume to be the north pole. Then stereographic projection is an embedding of the graph into the plane.

The general question also has a positive answer. It suffices to show that even if a combinatorial immersion covers the sphere, it is combinatorially equivalent to one which doesn't. Let $v \in V$. If $V$ combinatorially immerses in the sphere we have that $E(v)$ has cardinality of the continuum at most, where $E(v)$ is the set of edges containing $v$ as a coordinate. Thus $E(v)$ can be combinatorially immersed in the Cantor Fan. If $G$ is combinatorially immersed in the sphere, then consider the sphere with a closed disc removed. This is homeomorphic to the sphere punctured by $v$. Then we can just draw in the Cantor Fan in this excised neighborhood connecting as needed and obtain that in at least a neighborhood of one point of $G$ we get that the combinatorially equivalent image of the immersion is at most one-dimensional. There is a minor issue about winding toward a circle vs. a point, but all rays converging onto $v$ will have to have the same winding and can be 'unscrewed' simultaneously. I will leave the details for that bit. The result follows.

Note however that this does not necessarily give you a continuous map of the ends of the Cantor Fan onto the boundary circle of the excised neighborhood. In fact it will be impossible to get a continuous function in some cases, since continuous maps from the Cantor set onto the circle are necessarily at least $2$-to-$1$. So you will have to do your immersion in the fan in a potentially complicated and delicate way.

Edit one more time: Actually, I worry about using the Cantor Fan, for pedagogic reasons. Let us instead use a disc with a half-open crescent removed, one corner at $v$ and one at the boundary. Then we get a foliation that bijects with the boundary circle of the excised neighborhood in a trivial way. I am leaving up my first bad idea to give a new question: What conditions ensure that there always exist a vertex where the Cantor Fan is sufficient and can be immersed, as above, but continuously? I wonder, because if not you will have vertices all of whose neighborhoods of arcs are very fat, and compactness might smush some of these 'neighborhoods' together in ways that force some points to behave trivially. Though you might think about a Wada Basin scenario where the smushing isn't helpful. If you allow multigraphs, then a foliation by great circles through two antipodes does not allow this construction.

Answer 2

First, some definitions. I will say that a map of two topological spaces is an ICM if it is injective and continuous. In contrast, a map $f: X\to Y$ between topological spaces is called an embedding if it is a homeomorphism to its image (equipped with subspace topology). It is clear that there are no surjective topological embeddings from graphs to $S^2$ (since the latter is not homeomorphic to a graph). What you are asking about are ICMs, not topological embeddings.

Theorem. If $G$ is a graph which admits an ICM $f: G\to S^2$, then $G$ also admits an ICM to ${\mathbb R}^2$.

Proof. I will identify $S^2$ with the Riemann sphere, ${\mathbb C}\cup \{\infty\}$, the 1-point compactification of the complex plane. If $G$ has no edges then, because it has an injective map to $S^2$, the vertex-set of $G$ has cardinality of at most continuum. Therefore, any bijection $G\to {\mathbb R}^2$ will be an ICM.

Thus, I will assume that $G$ contains at least one edge. It follows that $G$ contains a topological arc $I$, a subset homeomorphic to the interval $[0,1]$ (just take a proper closed subarc in any edge). Then $f|_I$ is a topological embedding (since $I$ is compact and $S^2$ is Hausdorff) and, thus, $A=f(I)$ is homeomorphic to $[0,1]$, i.e. is a Jordan arc. I will use a nontrivial result of 2-dimensional topology: For every Jordan arc $J\subset {\mathbb R}^2$ there is a homeomorphisms of ${\mathbb R}^2$ to itself sending $J$ to a straight-line segment. See for instance here and here.

Remark. For the proof it suffices to know less, namely that the complement of any Jordan arc in $S^2$ is simply-connected. However, assuming only this, I would have to work a bit harder than I like.

It follows that there is a a homeomorphism $h: S^2\to S^2$ sending $A=f(I)$ to the subset $$ J=[0,\infty]\subset {\mathbb C}\cup \{\infty\}, $$ the 1-point compactification of the (real) positive half-line in the complex plane. On the complement ${\mathbb C} - J$ we have a single-valued branch of the function $\sqrt{z}$: If $z=re^{i\phi}, 0<\phi<2\pi$, then $$ \sqrt{z}= \sqrt{r}e^{i\phi/2}. $$ This function extends continuously to $0$ and $\infty$ by $$ \sqrt{0}=0, \sqrt{\infty}=\infty. $$ Consider the map $$ \sqrt{h\circ f} : G \setminus I \to U= \{z: Im(z)>0\}. $$ Extend this map to the end-points $p, q$ of the arc $I$ (sending them, respectively, to $0, \infty$). This defines a continuous injective map $$ g: (G\setminus I) \cup \{p, q\}\to U \cup \{0, \infty\}. $$ Extend this map to the arc $I$ by the map $h\circ f$. I will leave it to you to check that this extension defines an ICM $$ g: G\to U\cup [0,\infty]. $$ Thus, we obtained an ICM $g: G\to S^2 \setminus \{-i\}$ (I could have taken any point in the lower half-pane instead of $-i$). Since $S^2 \setminus \{-i\}$ is homeomorphic to ${\mathbb C}$, theorem follows. qed