Prove that there is a bijection between the space of closed paths in X with base point $x_0$ and the set of continuous functions

Question

Let X be a topological space and $x_0 \in X $ a point. Prove that there is a bijection between the space of closed paths in X with base point $x_0$ and the set of continuous functions $f: S^1\to X$ such that $f (1) = x_0.$ I don’t know how I have to show that this function is bijective. Can you help me?

Answer 1

Hint: firstly prove that $S^1 \equiv [0,1]/\sim$ where $\sim$ is the equivalence relation that identifies $0$ with $1$. A closed loop

$$ f : [0,1] \to X $$

is compatible with $\sim$, meaning that if $x \sim y$ then $f(x) = f(y)$. Now use a result on quotient spaces to obtain a map from the $1$-sphere to $X$. Finally, prove this assignment yields the desired bijection.

Edit: suppose that you have an equivalence relation $R$ on a space $X$. We define the quotient space $X/ R$ as the set of equivalence classes of $R$ with the final topology induced by the canonical projection $q : X \to X/R$. We say $h : X\to Y$ is $R$-compatible if $xRy$ implies $h(x) = h(y)$ for each $x,y \in X$. Now,

Proposition. Let $R$ be a equivalence relation on $X$ and $h : X\to Y$ an $R$-compatible continuous function. Then, there exists a unique continuous function $\widehat{h}: X/R \to Y$ such that $\widehat{h}q = h$.

Proof. Let's prove existence first. We define $\widehat{h}([x]) = h(x)$. Note that this is well defined (meaning it does not depend on the representative chosen) precisely because $h$ is $R$-compatible: if $[x] = [y]$ then $xRy$ and so $h(x) = h(y)$.

Recalling that $[x] = q(x)$, we see that moreover $\widehat{h}q = h$. This also proves continuity, because $X/R$ has the final topology with respect to $q$ and $\widehat{h}q$ is continuous by hypothesis.

As for uniqueness, this comes from the fact that $q$ is surjective and therefore right-cancellative: if $gq = h = \widehat{h}q$, then $g = \widehat{h}$. $\ \square$

Back to the original problem: if $f$ is $\sim$-compatible, we have a map $\widehat{f} : [0,1]/\sim \ \longrightarrow \ X$ as in the previous proposition.

Edit 2: Note that $[0,1]/\sim \equiv S^1$ via $g([t]) := e^{ti2\pi}$. We have seen that for each closed loop $f$ we can build a map from $[0,1]/\sim$ to $X$, which I've called $\widehat{f}$. Thus, we can define

$$ \begin{align} A &= \{\gamma : [0,1] \to X : \text{$\gamma$ is a closed loop based on $x_0$}\} \text{ and }\\ B &= \{f :S^1 \to X : \text{$f$ continuous and $f(1) = x_0$ }\} \end{align} $$

and

$$ \begin{align} F : \ &A \to B\\ &f \mapsto \widetilde{f} \end{align} $$

where $\widetilde{f} = \widehat{f}g^{-1}$.

(The intuition here is that we pass from a map on $[0,1]$ to one in the interval with glued endpoints, and then $g$ makes this into a map to $S^1$)

Let's see that $F$ is bijective: if $h : S^1 \to X$ is a map from the sphere to $X$, then $hgq$ is a map from $[0,1]$ to $X$. Moreover, by the uniqueness of the universal property it must be $\widehat{hgq} = hg$ and so $F(hgq) = \widetilde{hgq} = hgh^{-1} = h$. This shows that $F$ is surjective.

Finally, if $F(f) = \widehat{f}g^{-1} = \widehat{f'}g^{-1} = F(f')$ then $\widehat{f} = \widehat{f'}$ and so

$$ f = \widehat{f}q = \widehat{f'}q = f'. $$

To unpack all this, $g$ maps the interval with the glued ends to the circle, and $\widetilde{f}$ is just the mapping that takes a point $x \in S^1$, which is some point of the form $e^{ti2\pi}$, and outputs $f(t)$.

In a way, the idea here is that $S^1$ is like $[0,1]$ as long as we mentally glue the endpoints. Thus, a map on the circle can be 'flattened' to a map on $[0,1]$ which takes the same value on the $0$ and $1$. Reciprocally, every map from the interval that takes the same value at the endpoints can be made into a map from the circle by 'wrapping $[0,1]$ into $S^1$ (which makes the endpoints overlap, motivating this sort of coherence condition that we've required)'.

Answer 2

If $p:[0,1]\to X$ is a closed path with base point $x_0$, then $F(p): S^1 \to X$ can be defined by $F(p)(e^{2\pi it})= p(t)$, where $t \in [0,1)$.

$p \to F(p)$ is a bijection between your sets. If $p \neq p'$ they differ on some $t' \in [0,1]$ and then $F(p)$ and $F(p')$ differ on $e^{2\pi it'}$ and so are different loops.

And if $q: S^1 \to X$ is a loop with $q(1) = x_0$, just define $p(t) = q(e^{2\pi it})$ and check that $p$ is well-defined and $F(p) =q$, so we have ontoness.