# In a topological space if there exists a loop that cannot be contracted to a point does there exist a simple loop that cannot be contracted also?

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 $$0. Then $$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?

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 $$\mathbf{R}^3$$.

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.

• Thanks Anton Petrunin for the picture!
– YCor
Aug 8, 2019 at 21:56
• Thank you for this excellent example. I have a question which is that I can see why any simple loop must be on one cone and can then be contracted, but why can't non-simple loops that presumably must pass through w have their loops dropped down and passed over their respective cone points and then moved back up to w (I'm assuming w is the point where all the circles meet) and hence contract to a point. Aug 8, 2019 at 22:14
• 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 $1$.
– YCor
Aug 9, 2019 at 7:53
• 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 $\prod_na_nb_n$).
– YCor
Aug 9, 2019 at 13:11
• 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.
– 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.

• 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.
– YCor
Aug 9, 2019 at 3:49
• PS $\{0,\infty\}$ is missing among closed subsets in the example of my previous comment.
– YCor
Aug 9, 2019 at 14:59
• @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? 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.

• Nice example (simpler than mine); I added a picture.
– YCor
Aug 11, 2019 at 16:20
• 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. Aug 11, 2019 at 16:37
• @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.
– YCor
Aug 11, 2019 at 16:48
• @YCor Thanks for adding the picture.
– bof
Aug 11, 2019 at 17:39
• @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 $C$.
– bof
Aug 16, 2019 at 18:28