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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?

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    $\begingroup$ 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 $x'$. $\endgroup$ May 3 at 22:23

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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.

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