Question on showing a bijection between $\pi_1(X,x_0)$ and $[S^1, X]$ when X is path connected.

I am trying to do this question taken from Hatchers algebraic topology and I am struggling to understand the notation and the concepts.

As far as I know $\pi_1(X,x_0)$ is the set of end point preserving homotopy classes of loops in X based at $x_0$ and a loop is just a path $f:I \rightarrow X$ with $f(0)=f(1)$. The question says we can regard $\pi_1(X, x_0)$ as the set of basepoint preserving homotopy classes of maps $(S^1, s_0) \rightarrow (X, x_0)$.

This confuses me, does it mean the set of basepoint preserving homotopy class on maps $g:S^1 \rightarrow X$ which map loops in $S^1$ based at $s_0$ to loops in X based at $x_0$?

Then how can $\pi_1(X,x_0)$ be regarded as this set?

If someone could explain this it would be really useful

A based map $(S^1,s_0) \to (X,x_0)$ is a map $f: S^1 \to X$ with $f(s_0) = x_0$. (This is what a based map is for any pair of pointed spaces.)

Your notion of loop is the same as the notion of a based map from $S^1$. Note that $I/(0 \sim 1) \cong S^1$ as a topological space. By the definition of a quotient map, if you have a continuous map $f: I \to X$ such that $f(0)=f(1)=x_0$, this factors through the above quotient map, giving a map $f': I/(0 \sim 1) \to X$, with $f'([0]) = x_0$.

Conversely, given a based map $f: (S^1, s_0) \to (X,x_0)$ you can obtain a map $f': I \to X$, with $f'(0)=f'(1)=x_0$, essentially by undoing the above process.

Once you understand why these two notions of loop are actually (in the above sense) the same, it should not be difficult to go one step further and see why you can consider $\pi_1$ as (based) homotopy classes of (based) maps from $S^1$.