# Abelian fundamental groups

1. Jul 20, 2013

### R136a1

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

$$\Phi_\gamma : \pi_1(X,x)\rightarrow \pi_1(X,y): [f]\rightarrow [\overline{\gamma}]\cdot [f]\cdot [\gamma]$$

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. Jul 20, 2013

### WannabeNewton

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.