Image in freely homotopic group and conjugacy Let $f: \pi_1 (X, x_0) \rightarrow [S^1, X]$. Then $f([g]) = f([h])$ iff [g] and[h] are conjugate in $\pi_1(X, x_0)$. Could anyone give some suggestion on how I can go about proving this? I think what is holding me back is that I'm a bit lost as to what $f$ is exactly.
 A: $f$ is a function.
The input of $f$ is an element $[g] \in \pi_1(X,x_0)$.
To determine the output $f([g])$, recall that $g : [0,1] \to X$ is a path, i.e. a continuous function, with $g(0)=g(1)=x_0$. Also, $[g]$ is the path homotopy class of $g$, meaning the homotopy class relative to the endpoints. The output $f([g])$ is an element of $[S^1,X]$, which is the set of homotopy classes of continuous functions $S^1 \mapsto X$. So, using $g$, one must write a formula for a continuous function $G : S^1 \mapsto X$. This formula will do, once one checks that it is well defined when $t=0,1$:
$$G(e^{2 \pi i t}) = g(t)
$$
Once you've done that, $f([g]) = [G] \in [S^1,X]$ is the homotopy class of the function $G$. Now you have to prove that this is a well-defined formula, by showing that if $[g]=[h]$, namely if $g,h : [0,1] \to X$ have endpoints at $x_0$ and are path homotopic, then $[G]=[H]$, namely $G,H : S^1 \to X$ are homotopic.
Of course, you still have to prove what it asks you to prove, namely: given two closed paths $g,h : [0,1] \to X$, the equation $[G]=[H]$ holds if and only $[g]$ and $[h]$ are conjugate elements of the group $\pi_1(X,x_0)$, which by definition means that there exists $[k] \in \pi_1(X,x_0)$ such that $[k] [g] [k]^{-1} = [f]$, which translates to the statement that there exists a closed path $k : [0,1] \to X$ with endpoints at $x_0$ such that the concatenation $k * g * \bar k$ is path homotopic to $f$.
