(I'm trying to ask this in a way such that it isn't a duplicate question)
The proofs that I've seen for the fact that there is no retraction from the Mobius band to its boundary circle usually say that the homomorphism induced by inclusion is multiplication by 2, or they contradict the fact that the induced homomorphism from the retraction is surjective.
My problem is, at some point they talk about the fundamental group of the core circle as it sits inside of the Mobius band and do so ostensibly without worrying about basepoints. The basic idea is that one loop "wraps twice around another" - but the loops have different basepoints.
I guess the question then is, taking $M$ to be the mobius band, $\partial M$ to be its boundary and assuming we have a retraction $r$ gets us $r\circ i=1_{\partial M}$ where $i$ is inclusion. To account for basepoints, write $r:(M,x_0)\to (\partial M, y_0)$ and $i:(\partial M,x_0)\to (M, x_0)$; how do you write this up? Presumably you would trace a generator $[g]$ of $\pi_1(\partial M, x_0)$ through the commuting triangle given by $(1_{\partial M})_*=(r\circ i)_*$? I know that $i_*([g])$ can be explicitly computed, and there is a deformation retraction from $M$ to its core circle $C$.