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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 $mathbbR^n$?

I have thought about the simplest non-trivial case which I believe would be a subset of $mathbbR^2$.

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.

Here is an example of topological space $X$, embeddable as compact subspace of $mathbfR^3$, 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 $C=Htimes [0,1]/Htimes0$. 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 $mathbfR^3$.

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 $win X$ separates $Xsmallsetminusw$ 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.

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.