Bijection induced by mapping loops to their circle representations

Question

Let $X$ be a path connected space, and $p \in X$. I wish to show that the map $f \mapsto \tilde{f}$ induces a bijection between the conjugacy classes of $\pi(X,p)$ and $[\mathbb{S}^1: X]$, the free homotopy classes of continuous maps $\mathbb{S}^1 \to X$. Here, $f$ is a loop with base-point $p$ and $\tilde{f}$ is its unique circle representation. My initial guess is to construct the map sending a conjugacy class $\mathcal{C}$ to the circle representation of its representative, then sending that circle representation to its free homotopy equivalence class. Explicitly, $$ \mathcal{C} = [g] \mapsto \tilde{g} \mapsto [\tilde{g}] \in [ \mathbb{S}^1: X].$$ However, I'm not sure that this map is onto since a generic element in $[\mathbb{S}^1: X]$ need not even be a loop. I have thought about remedying this by the following (but I am not sure if there are errors): Take $\tilde{f} \in [\mathbb{S}^1: X]$, define the map $f := \tilde{f} \circ \omega$, where $\omega(t) = e^{2\pi i t}$. Since $X$ is path connected we can always form a new map $f'$ from $f$ that is a loop at $p$ (gluing new paths if necessary). However, I am not sure how to "put" this $f'$ into the correct conjugacy class. By this I mean that I'm not sure how to explicitly write down a pre-image of $\tilde{f}$. Am I on the right track? Any hints would be appreciated.

Answer 1

Let me write $\varphi$ for your map taking $[g]$ to $[\tilde{g}]$. There are three things you have to check to show $\varphi$ is a bijection:

1. $\varphi$ is well-defined: if $[g]=[g']$, then $[\tilde{g}]=[\tilde{g}']$.

2. $\varphi$ is injective: if $[\tilde{g}]=[\tilde{g}']$, then $[g]=[g']$.

3. $\varphi$ is surjective: if $[f]\in[\mathbb{S}^1:X]$, then there exists $g\in \pi_1(X,p)$ such that $[f]=[\tilde{g}]$.

As discussed in the question and in the comments, to prove (3), you note that $f$ can be viewed as a loop in $X$ based at some point $a$, and then concatenate it with a path between $a$ and $p$ on both sides to get a loop $g$ based at $p$. You then have to check that $\tilde{g}:\mathbb{S}^1\to X$ is homotopic to your map $f$.

Here is an indication of how you prove (2). If $[\tilde{g}]=[\tilde{g}']$, this means $\tilde{g}$ is homotopic to $\tilde{g}'$, and this homotopy can be viewed as a homotopy $H$ from $g$ to $g'$. However, this latter homotopy does not show $g$ and $g'$ are equal as elements of $\pi_1(X,p)$, because the endpoints of $g$ and $g'$ may not be kept fixed by it. However, we do know that at every stage of the homotopy, we have a map $\mathbb{S}^1\to X$, i.e. a loop, so the two endpoints of the path are still equal (they just might not be equal to $p$). Let $h$ be the path in $X$ given by following how the endpoints move under $H$ (that is, $h(t)=H(0,t)=H(1,t)$ for all $t$). Then $h$ starts and ends at $p$, since the basepoints of the loops $g$ and $g'$ are both $p$. So we can think about $h$ as an element of $\pi_1(X,p)$. What you want to prove now is that $g=hgh^{-1}$, as elements of $\pi_1(X,p)$. To show this, try and reparamaterize the homotopy $H$ to get a homotopy between $g$ and the concatenation $hgh^{-1}$ (if the domain of $H$ is a square $[0,1]\times[0,1]$, $g$ corresponds to the bottom edge, and $hgh^{-1}$ corresponds to going up the left edge, across the top edge, and then down the right edge).

The proof of (1) is similar to (2): you more or less reverse all the steps in the discussion above.