I'm interested in whether one only needs to consider simple loops when proving results about simply connected spaces.

If it is true that:

In a Topological Space, if there exists a loop that cannot be contracted to a point then there exists a simple loop that cannot also be contracted to a point.

then we can replace a loop by a simple loop in the definition of simply connected.

If this theorem is not true for all spaces, then perhaps it is true for Hausdorff spaces or metric spaces or a subset of $\mathbb{R}^n$?

I have thought about the simplest non-trivial case which I believe would be a subset of $\mathbb{R}^2$.

In this case I have a quite elementary way to approach this which is to see that you can contract a loop by shrinking its simple loops.

Take any loop, a continuous map, $f$, from $[0,1]$. Go round the loop from 0 until you find a self intersection at $x \in (0,1]$ say, with the previous loop arc, $f([0,x])$ at a point $f(y)$ where $0<y<x$. Then $L=f([y,x])$ is a simple loop. Contract $L$ to a point and then apply the same process to $(x,1]$, iterating until we reach $f(1)$. At each stage we contract a simple loop. Eventually after a countably infinite number of contractions we have contracted the entire loop. We can construct a single homotopy out of these homotopies by making them maps on $[1/2^i,1/2^{i+1}]$ consecutively which allows one to fit them all into the unit interval.

So if you can't contract a given non-simple loop to a point but can contract any simple loop we have a contradiction which I think proves my claim.

I'm not sure whether this same argument applied to more general spaces or whether it is in fact correct at all. I realise that non-simple loops can be phenomenally complex with highly non-smooth, fractal structure but I can't see an obvious reason why you can't do what I propose above.

Update: Just added another question related to this about classifying the spaces where this might hold - In which topological spaces does the existence of a loop not contractible to a point imply there is a non-contractible simple loop also?


3 Answers 3


Here is an example of topological space $X$, embeddable as compact subspace of $\mathbf{R}^3$, that is not simply connected, but in which every simple loop is homotopic to a constant loop.

Namely, start from the Hawaiian earring $H$, with its singular point $w$. Let $C$ be the cone over $H$, namely $C=H\times [0,1]/H\times\{0\}$. Let $w$ be the image of $(w,1)$ in $C$. Finally, $X$ is the bouquet of two copies of $(C,w)$; this is a path-connected, locally path-connected, compact space, embeddable into $\mathbf{R}^3$.

enter image description here

It is classical that $X$ is not simply connected: this is an example of failure of a too naive version of van Kampen's theorem.

However every simple loop in $X$ is homotopic to a constant loop. Indeed, since the joining point $w\in X$ separates $X\smallsetminus\{w\}$ into two components, such a loop cannot pass through $w$ and hence is included in one of these two components, hence one of the two copies of the cone $C$, in which it can clearly be homotoped to the sharp point of the cone.

  • 22
    $\begingroup$ Thanks Anton Petrunin for the picture! $\endgroup$
    – YCor
    Aug 8, 2019 at 21:56
  • 1
    $\begingroup$ Thank you again for your clarification. In trying to understand why my simple proof doesn't work in this case I have another question if that's OK? Can any $a_n$ or $b_n$ be contracted to w? (It seems intuitively that they can by sliding the loop down and around the respective cone's apex and back up to w). If they all can be then can't we just construct a homotopy out of these that contracts the entire infinite product to w? (If they can't then can we construct a counterexample out of two circles rather than an infinite number?) $\endgroup$
    – Ivan Meir
    Aug 9, 2019 at 7:31
  • 8
    $\begingroup$ Yes, each $a_n$ and each $b_n$ (and each finite product of these) is homotopic to a constant loop. A word is needed on this infinite loop: it is indexed by $[0,1]$, say $a_1$ indexed by $[0,1/2]$, $b_1$ by $[1/2,3/4]$, $a_2$ by $[3/4,7/8]$, etc. That it is well defined relies on the fact that $a_n$ and $b_n$ tend to $w$ uniformly. But if you concatenate homotopies, the concatenation is not continuous at time $1$. $\endgroup$
    – YCor
    Aug 9, 2019 at 7:53
  • 2
    $\begingroup$ Just to insist, the path defined as concatenation $\left(\prod_n a_n\right)\left(\prod_n b_n\right)$ is homotopic to a constant path (while it has the same support as $\prod_na_nb_n$). $\endgroup$
    – YCor
    Aug 9, 2019 at 13:11
  • 8
    $\begingroup$ PS this space is mentioned in MathSE post; it seems that the original reference is: H.B. Griffiths, The fundamental group of two spaces with a common point, Quart. J. Math. Oxford (2) 5 (1954) 175-190. $\endgroup$
    – YCor
    Aug 9, 2019 at 14:42

Every finite simplicial complex is weakly homotopy equivalent to a finite space. Therefore there are finite spaces with nontrivial loops; and these are obviously not embedded.

  • 4
    $\begingroup$ Nice non-Hausdorff examples! An explicit one (of the minimal possible cardinal) is the 4-element set $\{0,+,\infty,-\}$ with topology given by closed subsets $$\big\{\emptyset,\{0\},\{\infty\},\{0,+,\infty\},\{0,-,\infty\},\{0,+,\infty,-\} \big\}.$$ It can be view as quotient of the circle $\mathbf{R}\cup\{\infty\}$ by the action of the group of positive homotheties. Actually a lower-dimensional analogue of the question is whether there are spaces that are path-connected and not (injective path)-connected, and in this case the only examples are non-Hausdorff. $\endgroup$
    – YCor
    Aug 9, 2019 at 3:49
  • 3
    $\begingroup$ PS $\{0,\infty\}$ is missing among closed subsets in the example of my previous comment. $\endgroup$
    – YCor
    Aug 9, 2019 at 14:59
  • 2
    $\begingroup$ @YCor This 4-point space is apparently called the pseudocircle. (Not sure by whom?? Barmak-Minian have it as the minimal finite model of $S^1$, May as the non-Hausdorff suspension of $S^0$, denoted $\mathbb SS^0$.) But does it, or another like it, help answer the OP’s question? $\endgroup$ Aug 12, 2019 at 12:30

This question came up when I was taking a course in topology in a bygone century. For homework I constructed an example of a subspace $X$ of $\mathbb R^3$ which is not simply connected although every simple closed curve in $X$ is homotopic to a point. It was something like this:

Take an infinite sequence of circles in the $xy$-plane, each circle externally tangent to the next one, with the centers of the circles lying on a straight line and converging to the origin. For concreteness, we may suppose that the $n^\text{th}$ circle is a circle of radius $\frac1{2^n}$ centered at $\left(\frac3{2^n},0\right)$. Make each of those circles the base of a right circular cone of height $1$. Finally, let $X$ be the closure of the union of that sequence of cones. Every simple closed curve in $X$ can be shrunk to a point in $X$, since it lies on one cone; but a closed curve which goes around the bases of all the cones cannot be shrunk to a point in $X$.


From the same course I vaguely recall a proposition to the effect that, if $X$ is "locally simply connected in the large" (meaning that each point has a neighborhood $U$ such that every closed curve in $U$ is homotopic to a point in $X$), and if every simple closed curve in $X$ is homotopic to a point, then $X$ is simply connected. I don't recall if there were other conditions on $X$ (such as "Hausdorff space" or "metric space"), and I certainly don't recall anything about the proof, except that it could not have been anything deep.

P.S. This example seems to be well-known folklore. It is given without attribution in F. Galvin, Covering spaces and simple connectedness, M.A. thesis, University of Minnesota, 1961.

  • 3
    $\begingroup$ Nice example (simpler than mine); I added a picture. $\endgroup$
    – YCor
    Aug 11, 2019 at 16:20
  • 2
    $\begingroup$ This example is homotopy equivalent to the analogous compactification of the harmonic archipelago. However, the harmonic archipelago does have a non-contractible simple closed curve. This means the property the OP is interested in is not an invariant of homotopy type. $\endgroup$ Aug 11, 2019 at 16:37
  • $\begingroup$ @JeremyBrazas it's quite clear that for every path-connected not simply connected $X$, the space $X\times [0,1]^2$ has an injective non-contractible path (just make a homotopically non-trivial path in the $X$-direction and an injective path in the $[0,1]^2$-direction. $\endgroup$
    – YCor
    Aug 11, 2019 at 16:48
  • 2
    $\begingroup$ @AviSteiner Let $A=\{(x,y,z)\in X:z=0,y\ge0\}$, $B=\{(x,y,z)\in X:z=0,y\le0\}$, $C=\{(x,y,z)\in X:z=0\}$. Now $A$ is path-connected because it's homeomorphic to the closed interval $[0,2]$ via the homeomorphism $(x,y,z)\mapsto x$; $B$ is path-connected for the same reason; $C$ is path-connected because $C=A\cup B$ and $A\cap B\ne\emptyset$; and $X$ is path-connected because each point in $X$ is connected by a path in $X$ (a straight line segment) to a point in $C$. $\endgroup$
    – bof
    Aug 16, 2019 at 18:28
  • 2
    $\begingroup$ This other post mathoverflow.net/questions/220709/… also gives the same example. $\endgroup$
    – D.R.
    Aug 9, 2023 at 6:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.