Path space vs loop space

Question

If we consider the path fibration over a topological space $X$ we have $$ \Omega(X;p,p) \hookrightarrow P(X) \to X .$$ Where I denote with $\Omega(X;p,p)$ the set of paths $\omega:[0,1] \to X$ such that $\omega(0)=\omega(1)=p$. Why if we suppose that $X=M$ a Riemannian manifold, the space $\Omega(M,p,p)$ is homotopically equivalent to path space $\Omega(M;p,q)$ where $q \in M$ is not necessary equal to p?

Answer 1

I'll assume $M$ is connected so that there is a path from $p$ to $q$; if this assumption isn't satisfied then the assertion to be proved is false. So fix a path $\alpha$ from $p$ to $q$. Map $\Omega(p,p)$ to $\Omega(p,q)$ by sending each path $\gamma\in\Omega(p,p)$ to the concatenation $\gamma\alpha$; in the opposite direction, map $\Omega(p,q)$ to $\Omega(p,p)$ by sending each $\beta\in\Omega(p,q)$ to $\beta\alpha^{-1}$ (where $\alpha^{-1}$ is $\alpha$ traversed backward). The composition of these two transformations, in either order, is homotopic to the identity, because both $\alpha\alpha^{-1}$ and $\alpha^{-1}\alpha$ are homotopic to constant paths.