Conjugacy classes in the fundamental group I want to solve the following problem (Hatcher Ch.1, problem 6):

We can regard $π_1(X,x_0)$ as the set of basepoint-preserving homotopy classes of
  maps $(S_1, s_0)→(X,x_0)$. Let $[S_1,X]$ be the set of homotopy classes of maps $S_1→X$,
  with no conditions on basepoints. Thus there is a natural map $Φ :π_1(X,x_0)→[S_1,X]$
  obtained by ignoring basepoints. Show that $Φ $ is onto if $X$ is path-connected, and that
  $Φ([f]) = Φ([g])$ iff $[f]$ and $[g]$ are conjugate in $π_1(X,x_0)$. Hence $Φ$ induces a one-to-one correspondence between $[S_1,X]$ and the set of conjugacy classes in $π_1(X)$,
  when $X$ is path-connected.

To show that $Φ$ is onto, let $[\phi]$ be some element of $[S_1,X]$. Then it can be represented by some path $f$ at a point $x_1 \in X$. By path-connectedness, there is a path $\gamma$ connecting $x_0$ and $x_1$, so we can consider the path $\gamma \star f \star \bar{\gamma}$ based at $x_0$. Then there is a homotopy between $\gamma \star f \star \bar{\gamma}$ and $f$ (not base-point preserving) by continuously moving the basepoint from $x_0$ to $x_1$ through the path $\gamma$. Hence $\Phi[\gamma \star f \star \bar{\gamma}] = \Phi[f] = [\phi]$. 
However, I have no clue what to do to show the conjugacy part. 
 A: Sometimes it is easier to prove a more general result. Let $Y$ be a space with well pointed base point $y$, i.e. $(Y, \{ y \})$ has the HEP,  and let $X$ be a space with base point $x$. 
Consider the map of homotopy classes $p: [Y,X] _\bullet \to [Y,X]$ where the former is the base point preserving homotopy classes and the latter is the free homotopy classes. The result is that if $X$ is path connected then  $p$ is surjective and the group $\pi_1(X,x)$ operates on the set $[Y,X]_\bullet$ so that   the quotient is $[Y,X]$. 
This is actually 7.2.12 of Topology and Groupoids; the proof there uses the notion of fibration of groupoids, which is fun anyway, but the key hint is that you need the HEP to get the operation given above. I hope that helps. 
A: Use Lemma 1.19 in that very book. It says that if $\varphi_t:S^1\to X$ is a homotopy and if $h$ denotes the path $\varphi_t(s_0)$ formed by the images of $s_0$ then $\varphi_{1*}$ is equal to the composition 
$$\pi(S^1,s_0)\xrightarrow{φ_{0*}}\pi(X,φ_0(s_0))\xrightarrow{\beta_h}\pi(X,φ_1(s_0))$$
where $\beta_h([f])=[\overline h\cdot f\cdot h]$. In the case $φ_0(s_0)=φ_1(s_0)=x_0$ the path $h$ becomes a loop. Now given loops $f$ and $g$ at $x_0$, the equality $\Phi([f])=\Phi([g])$ implies a free homotopy $φ_t$ from $φ_0=f$ to $φ_1=g$ such that $φ_t(s_0)$ is a loop $h$. The formula for $\beta_h$ then gives the conjugacy. The other direction should not pose difficulties considering that you knew how to show the surjectivity.
