What exactly is the proof of onto here?

Question

I have been reading the solution of the following question:

We can regard $\pi_1(X, x_0)$ as the set of basepoint-preserving homotopy classes of maps $(S^1,s_0) \to (X, x_0).$ Let $[S^1, X]$ be the set of homotopy classes of maps $S^1\to X,$ with no conditions on basepoints. Thus there is a natural map $\Phi: \pi_1(X, x_0) \to [S^1, X]$ obtained by ignoring basepoints. Show that $(a)$ $\Phi$ is onto if $X$ is path-connected,\ $(b)$ Show that $\Phi([f]) = \Phi([g])$ iff $[f]$ and $[g]$ are conjugate in $\pi_1(X, x_0).$

Here: Why is $\phi$ onto if $X$ is path connected in Hatcher's exercise 1.1.6?

I am not quite sure where exactly is the proof of onto in this solution, could someone clarify this to me please?(I want a proof without using Homotopy extension property please if possible)

Answer 1

The fact that $\Phi$ is onto is quite simple. All it means is that every loop $f:[0,1]\to X$ is (freely) homotopic to a loop based at $x_0$. If $X$ is path connected, then there is a path $\lambda:[0,1]\to X$ from $x_0$ to $f(0)=f(1)$. Then $f$ is freely homotopic to the path composition $\lambda * f*\lambda^{-1}$, which is a loop based at $x_0$. And this follows from general homotopic properties of path composition.