- #1
R136a1
- 343
- 53
Hello everybody!
So, I've learned that in a path connected space, all fundamental groups are isomorphic. Indeed, if ##\gamma## is a path from ##x## to ##y##, then we have an isomorphism of groups given by
[tex]\Phi_\gamma : \pi_1(X,x)\rightarrow \pi_1(X,y): [f]\rightarrow [\overline{\gamma}]\cdot [f]\cdot [\gamma][/tex]
A problem here is that there is no "canonical" isomorphism. This means that if we are given two distinct paths ##\gamma## and ##\gamma^\prime## from ##x## to ##y##, then the isomorphisms ##\Phi_{\gamma}## and ##\Phi_{\gamma^\prime}## don't need to be equal.
Now, I read a comment somewhere that the isomorphisms are canonical in the case that the groups are all abelian. How would I justify this? So, I guess I'm asking why if the fundamental groups ##\pi(X,x)## and ##\pi_1(X,y)## are abelian, then for any paths ##\gamma## and ##\gamma^\prime## from ##x## to ##y##, we have ##\Phi_\gamma = \Phi_{\gamma^\prime}##.
Finally, is the converse true as well?
So, I've learned that in a path connected space, all fundamental groups are isomorphic. Indeed, if ##\gamma## is a path from ##x## to ##y##, then we have an isomorphism of groups given by
[tex]\Phi_\gamma : \pi_1(X,x)\rightarrow \pi_1(X,y): [f]\rightarrow [\overline{\gamma}]\cdot [f]\cdot [\gamma][/tex]
A problem here is that there is no "canonical" isomorphism. This means that if we are given two distinct paths ##\gamma## and ##\gamma^\prime## from ##x## to ##y##, then the isomorphisms ##\Phi_{\gamma}## and ##\Phi_{\gamma^\prime}## don't need to be equal.
Now, I read a comment somewhere that the isomorphisms are canonical in the case that the groups are all abelian. How would I justify this? So, I guess I'm asking why if the fundamental groups ##\pi(X,x)## and ##\pi_1(X,y)## are abelian, then for any paths ##\gamma## and ##\gamma^\prime## from ##x## to ##y##, we have ##\Phi_\gamma = \Phi_{\gamma^\prime}##.
Finally, is the converse true as well?
Last edited: