1.1.6: 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 $\phi :π_1(X,x_0)→[S_1,X]$ obtained by ignoring basepoints. Show that $\phi$ is onto if $X$ is path-connected, and that $\phi([f ]) = \phi([g])$ iff $[f ]$ and $[g]$ are conjugate in $\phi_1(X,x_0)$. Hence $\phi$ induces a one to-one correspondence between $[S1,X]$ and the set of conjugacy classes in $π_1(X)$, when X is path-connected.
Suppose that $X$ is path connected, I want to show that $\phi$ is onto.
I don't understand why path connectedness is needed here. Take any $[f]\in \pi_1(X,x_0)$. $f:I\to X$ is a loop based at $x_0$. By universal property of quotient maps, there exists a unique $\tilde f: S^1\to X, \tilde f\circ q=f,$ where $q:I\to S^1$ is the quotient map.
So I define $\phi([f])= [\tilde f]$.
To show that $\phi$ is onto, let $[g] \in [S^1, X]$ be chosen arbitrarily. Then, we have as per the definition above $\phi([g\circ q])=[g]$ so $g$ is onto.
What is wrong with the above and how to fix it? Thanks.
Consider a space $X$ having more than one path-component. Let $P_0$ be the path-component of $x_0$. Then each map $f : (S^1,s_0) \to (X,x_0)$ has image in $P_0$. Given a map $F : S^1 \to X$ with image in a path-component $P \ne P_0$, it cannnot be homotopic to map with image in $P_0$. Thus $[F] \notin \phi(\pi_1(X,x_0))$.
If $X$ is path-connected, then $\phi$ is indeed onto. Given any map $F : S^1 \to X$, we have to find $f : (S^1,s_0) \to (X,x_0)$ such that $f \simeq F$. Le $x_1 = F(s_0)$ and $F' = F \circ q : I \to X$. Choose a path $u : I \to X$ such that $u(0) = x_0$ and $u(1) = x_1$. The loop $$F'' : I \to X, F''(s) = \begin{cases} x_1 & s \le 1/3 \\ F'(3s-1) & 1/3 \le s \le 2/3 \\ x_1 & 2/3 \le s \end{cases}$$ is homotopic rel. $\{0,1\}$ to the loop $F'$.
Define a homotopy $$H : I \times I \to X, H(s,t) = \begin{cases} u(t + 3s(1-t)) & s \le 1/3 \\ F'(3s-1) & 1/3 \le s \le 2/3 \\ u(t + 3(1-s)(1-t)) & 2/3 \le s \end{cases}$$ Then $H(0,t) = H(1,t) = u(t)$ for all $t \in I$ which means that $H_t = H(-,t) : I \to X$ is a loop based at $u(t)$. Moreover $H_1 = F''$.
The map $Q = q \times id_I : I \times I \to S^1 \times I$ is a quotient map and by construction $H$ is constant on all fibers $Q^{-1}(z,t)$ of $Q$ (which are singletons if $z \ne s_0$ or hav ethe form $\{0,1\} \times \{t\}$ for $z = s_0$). Thus $H$ induces a homotopy $h : S^1 \times I \to X$. Leeting $f = h_0$, we see that $f(s_0) = x_0$ and $f \simeq F$.