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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?

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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.

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    $\begingroup$ Thanks Anton Petrunin for the picture! $\endgroup$ – YCor Aug 8 at 21:56
  • $\begingroup$ 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. $\endgroup$ – Ivan Meir Aug 8 at 22:14
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    $\begingroup$ 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 at 7:53
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    $\begingroup$ 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 at 13:11
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    $\begingroup$ 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 at 14:42
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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.

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    $\begingroup$ 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 at 3:49
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    $\begingroup$ PS $\{0,\infty\}$ is missing among closed subsets in the example of my previous comment. $\endgroup$ – YCor Aug 9 at 14:59
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    $\begingroup$ @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? $\endgroup$ – Francois Ziegler Aug 12 at 12:30
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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 $X$.

picture

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.

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    $\begingroup$ Nice example (simpler than mine); I added a picture. $\endgroup$ – YCor Aug 11 at 16:20
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    $\begingroup$ 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. $\endgroup$ – Jeremy Brazas Aug 11 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 at 16:48
  • $\begingroup$ @YCor Thanks for adding the picture. $\endgroup$ – bof Aug 11 at 17:39
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    $\begingroup$ @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 at 18:28

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