Let $\Phi: \pi_1(X, x_0) \rightarrow [S^1, X]$. I wish to show that $\Phi([f]) = \Phi([g])$ iff $[f]$ and $[g]$ are conjugate in $\pi_1(X, x_0)$.
To me this this is almost trivial, since $\Phi([f]) = \Phi([g])$ gives us that we have a homotopy between $[f]$ and $[g]$, and since both are in the same fundamental group they both are loops around $x_0$. However, I know this is wrong and I am misunderstanding something. Furthermore, I know that the right approach is something along the lines of constructing $f = \overline{h}gh$, but why? All are loops centered around the same point, so what is this solving really? How is it "moving" the base point if all have the same base point? I have been beating my head against the wall for hours with this problem. Any help would be much appreciated.
Thank you.
Best Answer
If $\Phi([f]) = \Phi([g])$, let $F:I \times I \to X$ be a free homotopy of loops between $f$ and $g$ where the bottom edge of $I \times I$ is mapped to $f$ and the top edge to $g$. Let $h$ be the right edge of $I \times I$, i.e., the loop $h(t) = F(1, t)$. Note that $h^{-1}$ is exactly the loop $F(0, 1-t)$ (the left edge traversed downwards) because $F$ is a homotopy of loops so $F(0, t) = F(1, t)$ for all $t \in I$. The boundary loop of $I \times I$ traversed anti-clockwise is nullhomotopic in $I \times I$ while preserving the basepoint $(0, 0)$. This induces a basepoint-preserving nullhomotopy of $fhg^{-1}h^{-1}$, showing that $[f]$ and $[g]$ are conjugate in $\pi_1(X, x_0)$. Conversely if $[f] = [hgh^{-1}]$ let $F: I \times I \to X$ be a basepoint-preserving homotopy. Since the left and right edges of $I \times I$ are mapped to constant loops at $x_0$, we might as well consider them to be $h$ and $h^{-1}$. Then we are left with a free homotopy of loops between $f$ and $g$.