Can a simple, closed curve be homeomorphic to a closed curve that is not simple?

Question

It seems like most standard definitions of simple curves in $\mathbb{R^2}$ are those without any self-intersections. Moreover, it can be shown that if $C$ is a simple closed curve, then $C$ is homeomorphic to the circle $S^1$.

(All using the subspace topology of the standard Euclidean topology on $\mathbb{R^n}$ for both, that is.)

So suppose a curve $C$ is not simple, so that it has at least one point of self intersection. That is, if $C$ is identified as the image of a continuous function $f: S^1 \rightarrow \mathbb{R}^2$ from the circle to the plane, then there is at least one point $p\in S^1$ at which $f$ fails to be injective.

In this case, can $C$ still be homeomorphic to a circle?

My hunch is that it can not be, and here's what I've come up with so far:

Towards a direct proof, I think that if $q = f(p)$ is such a point of self-intersection on the curve $C$, then the inverse function $f^{-1}$ may fail to be continuous at $q$. Intuitively, it seems like neighborhoods around this point will always contain points from at least two "branches" of $C$, but I'm not sure how to proceed from here.

Using other topological invariants seemed promising - I think deleting the point $q$ breaks $C$ into at least two connected components while deleting its preimage $p$ leaves the circle in one connected component. Does this constitute a valid proof?

Edit: Phrased another way, if we identify knots as continuous images of $S^1$ embedded in $\mathbb{R^3}$, when is a knot homeomorphic to $S^1$?

Answer 1

You might be interested in the Hahn-Mazurkiewicz Theorem. It tells you that there are a lot of weird subsets of $\mathbb{R}^2$ which are "closed curves", or more properly speaking which are the images of a continuous function $f : S^1 \to \mathbb{R}^2$, for example the Sierpinski triangle.

In fact the theorem characterizes exactly which topological spaces (with a few conditions) can be continuous images of $S^1$. And, of course, if the space is not homeomorphic to $S^1$ then the map $f$ will not be injective (as the comment of @LordSharkTheUnknown shows, the converse is not true).

Here's the usual, very general statement of the theorem, using $[0,1]$ as the domain:

• A non-empty Hausdorff topological space is a continuous image of $[0,1]$ if and only if it is a compact, connected, locally connected metrizable space.

Since $[0,1]$ is the continuous image of $S^1$ and $S^1$ is the continuous image of $[0,1]$, we obtain another version of the theorem as an immediate corollary:

• A non-empty Hausdorff topological space is a continuous image of $S^1$ if and only if it is a compact, connected, locally connected metrizable space.

You ask only about subsets of the plane, so you can of course specialize this theorem to apply only to such subsets:

• A nonempty subset of $\mathbb{R}^2$ is a continuous image of $S^1$ if and only if it is compact, connected, and locally connected.

That's a lot of weird subsets.

The construction here shows visually that the Sierpinski Curve is a continuous image of $[0,1]$, and hence also of $S^1$.

Answer 2

Take$$\begin{array}{rccc}f\colon&S^1&\longrightarrow&\mathbb{R}^2\\&(x,y)&\mapsto&\left(\frac x{y^2+1},\frac{xy}{y^2+1}\right).\end{array}$$Then $f$ is not injective, because $f(0,1)=f(0,-1)=(0,0)$. And the image of $f$ is a Lemniscate of Bernoulli, which is not homeomorphic to $S^1$, because if we remove the point $(0,0)$ from it, the we get wo connected components. That does not occur with $S^1$.

Answer 3

Your notion of disconnecting points into separate components can also be made local - let's call them 'cut points.' We'll say a local cut point of order $k$ is a point $x$ with a neighborhood $U$ containing $x$ so that $U - \{x\}$ has $k$ path components.

The reason we need a local notion is because we can make simple closed curves for which deleting a point doesn't separate the curve. Take a money sign for example $, and close it by attaching the one end of the S to the other. If we remove a point of self-intersection, we can still go around the other way. But a homeomorphism will preserve local cut points as well as global cut points, and so the kind of invariant you want will work out in general, since local connectedness (and the number of path components of a deleted neighborhood) are both preserved by continuous functions, and hence homeomorphisms.