My question is pretty simple although I have not been able to find an answer yet:
Let $c_{x_0}\in\Omega(X,x_0)$ denote the constant loop in $x_0$. Then, by standard homotopy theoretical arguments, we have an isomorphism of abelian groups $\pi_1(\Omega(X,x_0),c_{x_0})\cong\pi_2(X,x_0)$. Now my question is when we choose a non-constant loop $\xi\in\Omega(X,x_0)$, which does not lie in the same path component as $c_{x_0}$ (assuming that $\Omega(X,x_0)$ consists of more than just one path component), does there still exist an isomorphism of the form $\pi_1(\Omega(X,x_0),\xi)\cong\pi_2(X,x_0)$? If yes, how does the proof look like and if not, are there counter examples?
Path components in $\Omega(X,x_0)$ are homotopy equivalent. Let $A$ be path component of $c_{x_0}$ and $B$ any path component. Fix $d\in B$. Then homotopy equivalence $A\to B$ is given by $p\mapsto p*d$ while its homotopy inverse $B\to A$ is given by $p\mapsto p*d^{-1}$, where "$*$" is path concatenation and $d^{-1}$ is the reversed loop.
That way $\pi_1(\Omega(X,x_0))$ does not depend on the choice of basepoint.
This is a special case of situation where we have continuous multiplication on the set of path components.