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Abelian fundamental groups

  1. Jul 20, 2013 #1
    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?
     
    Last edited: Jul 20, 2013
  2. jcsd
  3. Jul 20, 2013 #2

    WannabeNewton

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    Science Advisor

    Yes the converse is true. For the first direction, let ##\pi_{1}(X,x)## be abelian and ##g,h## two paths from ##x## to ##y##. Note that ##g\bar{h}## is a loop based at ##x## so ##[g][\bar{h}]\in \pi_{1}(X,x)##, where the bar denotes the reverse path. Now let ##[f]\in \pi_{1}(X,x)## then ##\Phi_{g}([f][g][\bar{h}]) = [\bar{g}][f][g][\bar{h}][g] = \Phi_{g}([g][\bar{h}][f]) = [\bar{h}][f][g]## thus ##[\bar{g}][f][g] = [\bar{h}][f][h]## hence ##\Phi_{g} = \Phi_{h}##.

    As for the converse, take ##x = y## and ##[f_1],[f_2],[c_{x}]\in \pi_{1}(X,x)## where ##c_{x}## is the constant loop based at ##x##. Then ##[\bar{f_2}][f_1][f_2] = [f_1]## so ##[f_1][f_2] = [f_2][f_1]## i.e. ##\pi(X,x)## is abelian.
     
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