I've recently read some papers and books involving simply connected domains in Euclidean space (dimension at least 2), where domain is an open connected set. The usual definition is a (connected) set for which every continuous closed curve is (freely) contractible while some authors only require that every continuous simple closed curve is contractible. The authors who define simple connectedness using simple closed curves do so in order to use Stokes' Theorem or the Jordan curve theorem somewhere in the sequel; however, they never mention (not even with a reference) that their definition is equivalent to the usual one! My question is if there is a proof written down somewhere (with all the details) proving the equivalence (for domains in $\mathbb{R}^n$ with $n \geq 2$)? If not, does someone know of an "easy" proof using a minimal amount of knowledge, say that of a first course in topology?

8$\begingroup$ Any curve in an open set of euclidean space is homotopic to a piecewiselinear curve (cover the curve with balls contained in the domain...). And in turn, any piecewiselinear curve is homotopic to a product of simple ones (product in the sense of the fundamental group). So if all simple curves are contractible, then all curves are contractible. Btw you may want to worry about basepoints, and whether the homotopies (in the book you are reading) are assumed to be through simple curves or not. $\endgroup$– PierreApr 5, 2011 at 20:59

3$\begingroup$ @Pierre: your comment seems like a correct answer to me. Would you be so kind as to leave it as such? $\endgroup$– Pete L. ClarkApr 6, 2011 at 2:55

1$\begingroup$ Topologists invoke "transversality" to such questions, which refers to a series of theorems that rigorously prove intuitive statements like "every continuous function from a circle to a smooth manifold of dimension > 2 is $\epsilon$homotopic to a smooth embedding", or "every continuous function from a circle to a smooth 2manifold is $\epsilon$homotopic to a finite union of simple closed curves." These theorems aren't trivial to prove, but they form the basis on which most statements like the one you ask follow. Careful expositions are found in e.g. in Hirsch's book. $\endgroup$– PaulApr 6, 2011 at 4:04

$\begingroup$ @Pierre: Is there a reference to prove that "any piecewiselinear curve is homotopic to a product of simple ones (product in the sense of the fundamental group)". Although this statement is intuitively clear, how does one prove it carefully? (Last week I've asked a famous topologist exactly this question and he showed me carefully how to do it  the proof is very long if one were to write it out in all details.) $\endgroup$– simply connectedApr 6, 2011 at 6:44
2 Answers
As pointed out by Pierre and Paul in comments, there are several standard ways to deal with this kind of issue. A good answer really depends what you're assuming you start from, and where you're trying to go to. The Jordan curve theorem and Stoke's theorem are both fairly sophisticated and difficult for beginners to grasp, so it's a bit hard to see how only analyzing embedded curves is streamlining anything, except perhaps helping with people's intuitive imagesbut even so, it may do more harm than good.
Perhaps it's worth pointing out that this statement is false in greater generality, for instance for closed subsets of $\mathbb R^3$. Here's an example in $\mathbb R^3$: consider a sequence of ellipsoids that get increasingly getting long and thin; to be specific, they can have axes of length $2^{k}$, $ 2^{k}$ and $2^k$. Stack them in $\mathbb R^3$ with short axes contained in the $z$axis, so each one touches the next in a single point with long axes parallel to the $x$axis, and let $X$ be their union together with the $x$axis.
Any simple closed curve in $X$ is contained in a single ellipsoid, since to go from one to the next it has to cross a single point, so every simple closed curve is contractible.
However, a closed curve in the $yz$crosssection that goes down one side and back up the other sides is not contractible. The fundamental group is in fact rather large and crazy.
Anyway, here are some lines of reasoning that can overcome whatever hurdle needs to be ovrcome:
PL approximation, as suggested by Pierre: this is easy, the keyword is "simplicial approximation". I'll phrase it for maps of a circle to Euclidean space as in the question, even though essentially the same construction works in far greater generality. Given an open subset $U \subset \mathbb{R}^n$ and given a map $f: S^1 \to U$, then by compactness $S^1$ has a finite cover by neighborhoods that are components of $f^{1}$ of a ball. If $U_i$ is a minimal cover of this form, there is a point $x_i$ that is in $U_i$ but not in any other of elements of the cover; this gives a circular ordering to the $U_i$. There is a sequence of points $y_i \in U_i \cap U_{i+1}$, indices taken mod the number of elements of the cover. The line segment between $y_i$ and $y_{i+1}$ is contained in $U_i$, since balls are convex. (This generalizes readily to the statement that for any simplicial complex, there is a subdivision where the extension that is affine on each simplex has image contained in $U$. It also generalizes readily to the case that $U$ is an open subset of a PL or differentiable manifold).
Raising the dimension: if you take the graph of a map of $S^1$ into a space $X$, it is an embedding. If you're (needlessly) worried about integrating differential forms on nonembedded curves, pull the forms back to the graph, where the curve is embedded. If you want to map to a subset of Euclidean space with the same homotopy type, just embed the graph of the map (a subset of $S^1 \times U$ into $\mathbb R^2 \times U$. (There's a very general technique to do this, if the domain is a manifold more complicated than $S^1$, even when it's just a topological manifold, using coordinate charts together with a partition of unity to embed the manifold in the product of its coordinate charts).
The actual issue for integration, using Stoke's theorem etc., is regularity  to make it simple, restrict to rectifiable curves, and don't worry about embededness. Any continuous map into Euclidean space is easily made homotopic to a smooth curve, by convolving with a smooth bump functionthe derivatives are computed by convolving with the variation of the bump, as you move from point to point.
Similarly, you can approximate any continuous map by a realanalytic function, if you convolve with a time $\epsilon$solution of the heat equation (a Gaussian with very small variance, wrapped around the circle). This remains in $U$ if $\epsilon$ is small enough. A real analytic map either has finitely many double points, or is a covering space to its image; in either case you reduce simple connectivity to the case of simple curves.
Sard's theorem and transversality, as mentioned by Paul. Sard's theorem is nice and elegant and has many applications, including the statement that a generic smooth map of a curve into the plane is an immersion with finitely many selfintersection points, as is any generic smooth map of an $n$manifold into a manifold of dimension $2n$. If the target dimension is greater than $2n$, then a generic smooth map is an embedding.

$\begingroup$ It took me a few minutes to figure this out, so in case this helps anyone else to "get" the picture in the second paragraph: a "short circle" means a circle like a belt on one of the (rotationally symmetric) ellipsoids with radius equal to the semiminor axis, say $2^{k}$. $\endgroup$– j.c.Apr 7, 2011 at 1:10

$\begingroup$ @jc7: Thanks for the comment. I hadn't anticipated this way my phrasing could mislead  I'll edit and see if I can bring forward the intended mental image. $\endgroup$ Apr 7, 2011 at 9:07
The equivalence is conceptually easy: each closed curve is a union of simple closed curves. If you can contract each simple closed curve, you can contract the whole curve. Each simple closed curve also lives in the set of closed curves, so the equivalence the other direction is simple. This sort of proof shouldn't be too hard for you to construct, assuming you have the knowledge of a first course in topology. Some care might need to be taken in constructing the explicit homotopy and in dealing with a curve which has infinitely many selfintersection points, but these are both issues you should have seen in such a first course in topology.

2$\begingroup$ This seems to fall short of rigorous. Suppose for instance that the image of the curve contains the unit cube in Euclidean space. How are you going to decompose it as a union of simple closed curves? $\endgroup$ Apr 6, 2011 at 2:14

$\begingroup$ @Pete L. Clark This answer is helpful, because a closed curve in an open set in $\mathbb{R^n}$ is homotopic to a piecewiselinear curve made of finitely many line segments (This is not hard to prove, with the aid of the paving lemma). Therefore, we can reduce this to the case where the curve is not a spacefilling curve, nor does it have infinitelymany selfintersection points. $\endgroup$ Sep 2, 2021 at 9:24