Isomorphism between fundamental groups. Can anyone give some advice to find the conditions to the following exercises:


*

*Under what conditions will two paths clasess, $\gamma$ and $\gamma'$, from $x$ to $y$ give rise to the same isomorphism of $\pi (X,x)$ onto $\pi(X,y)$? 

*Let $X$ be an arcwise-connected space. Under what contidions is the following statement true: For ant two points $x,y \in X$, all path classes from $x$ to $y$ give rise to the same isomorphism of $\pi(X,x)$ onto $\pi(X,y)$.
I know that the second follow from the first. 
Note: $\pi(X,x)$ is the fundamental group of $X$ at $x$. 
Thanks!
 A: Suposse that for $\gamma, \gamma'$ (connecting $x$ to $y$)  we have the same isomorphism then for every $\phi\in\pi(X,x)$ we have 
$$\gamma^{-1}\phi\gamma=\gamma'^{-1}\phi\gamma' $$
which implies $$\gamma'\gamma^{-1}\phi\gamma\gamma'^{-1}=\phi $$
Consequently $\gamma'\gamma^{-1}=(\gamma\gamma'^{-1})^{-1}$ and $\gamma'\gamma^{-1}\in \pi(X,x)$. Furthermore the equation above implies that $\gamma \gamma'^{-1}$ belongs to the center. And the converse is evidently.
We have got a criterion: $\gamma$ and $\gamma'$ rise the same isomorphism if and only if $\gamma\gamma'^{-1}$ belongs to the center of $\pi(X,x)$
A: It obviously suffices if $\gamma$ and $\gamma^\prime$ are homotopic rel.\ endpoints, i.e., there is a homotopy that keeps the endpoints fixed during the whole homotopy. 
To see that this is also a necessary condition: $\gamma^{-1}\gamma^\prime$ represents an element $g$ of $\Pi(X,y)$ and you are asking whether the action of $g$ by right multiplication on $\Pi(X,y)$ is the identity. But this is the case only if $g$ is the neutral element, which implies the above condition.
A: This is really a question about groupoids, see for example the downloadable book Categories and Groupoids, or this book Topology and Groupoids.  
If $G$ is a groupoid and $x \in Ob(G)$ we write $G(x)$ for the vertex group at $x$. If $a: x \to y$ in $G$ then we get an isomorphism $a_*: G(x) \to G(y)$ by conjugation by $a$. You can then show that $a_*= b_*$ for all $a,b: x \to y$ if and only if $G(x)$ is abelian. 
