# Can a simply-connected topological space be written as the union of the images of all possible loops at a given basepoint?

Can a simply-connected topological space be written as the union of the images of all possible loops at a given basepoint?

Let $$X$$ be a simply-connected topological space, fix $$x\in X$$, and let $$\{\gamma_i\}_{i\in I}$$ be the collection of possible loops $$\gamma: [0,1] \to X$$ with $$0\mapsto x$$ and $$1\mapsto x$$ (over some indexing set $$I$$).

Insofar as the concept of "collection of all possible loops" makes sense, is it accurate to say that $$X$$ can be represented as the union of the images of each of the loops in the collection?

Not sure how you would notate this, but maybe, $$\bigcup_{i \in I} \text{Im} \gamma_i$$ would work.

Geometrically this seems true, since you could "catch" any point in $$X$$ just by defining a loop that goes through that point, and since $$X$$ is simply connected, all the loops are homotopic (and nullhomotopic), so you can catch all the points with loops in the collection.

But, this is just my geometric intuition: are there any counterexamples to this, or issues with this line of argument?

• This is true for any path-connected space because for every point $x' \in X$, there is a loop based at $x$ which passes through $x'$. May 3 at 22:23

Lemma. Let $$X$$ be a topological space and $$x_0\in X$$ a point. Then $$X$$ is path connected if and only if for any $$x\in X$$ there is a loop at $$x_0$$ passing through $$x$$.

Proof. "$$\Leftarrow$$" an exercise.

"$$\Rightarrow$$" Since $$X$$ is path connected then there is a path $$\lambda:[0,1]\to X$$ such that $$\lambda(0)=x_0$$ and $$\lambda(1)=x$$. Then $$\lambda *\overline{\lambda}$$ is the loop we are looking for, where $$\overline{\lambda}(t)=\lambda(1-t)$$ and "$$*$$" stands for path composition. $$\Box$$

Given a topological space $$X$$ and a fixed point $$x_0\in X$$ write down

$$C([0,1],X,x_0)=\{\lambda:[0,1]\to X\ |\ \lambda(0)=\lambda(1)=x_0\text{ and }\lambda\text{ is continuous}\}$$

Then the lemma above leads quite easily to the following:

Corollary. A topological space $$X$$ is path connected if and only if $$X=\bigcup_{\lambda\in C([0,1],X,x_0)}\text{Im}\lambda$$

And so being simply connected is too strong, path connected is enough.