6

In the Munkres' book of topology have this question:

$ {\pi}_1 (X, x_0) $ is abelian if and only if for every pair $\alpha $ and $\beta $ of paths from $x_0$ to $x_1$, we have $ \widehat{\alpha} = \widehat{\beta} $. where $X$ is a path-connected space.

to prove in the direction $(\Rightarrow)$ I think of using, for example

$ \widehat{\alpha}([f]) = \widehat{\beta}([f]) $.

$\Leftrightarrow$

$[\bar{\alpha}]*[f]*[\alpha] = [\bar{\beta}]*[f]*[\beta] $

$\Leftrightarrow$

$([\bar{\alpha}]*[\beta])*([\bar{\beta}]*[f]*[\alpha]) = ([\bar{\beta}]*[f]*[\bar{\alpha}])*([\alpha]*[\beta]) $

but $[\bar{\alpha}]*[\beta]\notin {\pi}_1 (X, x_0) $ and $[\bar{\beta}]*[f]*[\alpha] \notin {\pi}_1 (X, x_0) $

and thus I can not use commutativity to complete the proff. I tried other ways to arrange the terms but I always fall into the same problem.

still, I can not carry out the operation $[\alpha]*[\beta]$ because $\alpha(1) \neq \beta(0)$.

Luiz Collovini
  • 357
  • This exercise is on page 335, section fundamental group. – Luiz Collovini Nov 17 '17 at 10:42
  • 1
    Notice that $\pi_1(X,x_0)\cong \pi_1(X,x_1)$ if $x_0$ and $x_1$ belong to the same path component. Thus assuming $\pi_1(X,x_0)$ to be abelian also yields that $\pi_1(X,x_1)$ is abelian. – Mathematician 42 Nov 17 '17 at 12:02
  • Very good your idea, I did not realize that I could use this. but on the right side I can not do the operation $[\alpha]*[\beta]$ because $\alpha(1) \neq \beta(0)$. when I try to get around that problem, a problem arises elsewhere. – Luiz Collovini Nov 17 '17 at 12:17
  • 1
    Your last equivalence is wrong, you can never write $[\alpha]*[\beta]$ as this operation is not defined. (one of those paths should be reversed). – Mathematician 42 Nov 17 '17 at 12:23

2 Answers2

11

Suppose $\pi_1(X,x_0)$ is abelian (and thus $\pi_1(X,x_1)$ as well).

Then \begin{eqnarray} [\overline{\alpha}]*[f]*[\alpha] &=& [\overline{\alpha}]*[f]*([\beta]*[\overline{\beta}])*[\alpha]\\ &=& ([\overline{\alpha}]*[f]*[\beta])*([\overline{\beta}]*[\alpha])\\ &=& ([\overline{\beta}]*[\alpha])*([\overline{\alpha}]*[f]*[\beta])\\ &=& ([\overline{\beta}]*[f]*[\beta]). \end{eqnarray}

Here I used that $([\overline{\alpha}]*[f]*[\beta])$ is a loop at $x_1$ and $([\overline{\beta}]*[\alpha])$ is also a loop at $x_1$.

EDIT: as requested in the comments, we will prove the reverse implication as well.

Let $f$ and $g$ be two loops with base point $x_0$ and let $\alpha$ be a path from $x_0$ to $x_1$. Then $\beta=f*\alpha$ is a path from $x_0$ to $x_1$ as well. By assumption, $\widehat{\alpha}(g)=\widehat{\beta}(g)$. Hence \begin{eqnarray} [\overline{\alpha}]*[g]*[\alpha] &=& [\overline{\beta}]*[g]*[\beta]\\ [\beta]*[\overline{\alpha}]*[g]*[\alpha] &=& [g]*[\beta]\\ [f]*[g]*[\alpha] &=& [g]*[\beta]\\ [f]*[g] &=& [g]*[\beta]*[\overline{\alpha}]\\ [f]*[g] &=& [g]*[f]\\ \end{eqnarray} This completes the proof.

Mathematician 42
  • 12,591
3

Apart from answer given by Mathematician 42, if you want to work within $\pi_1(X, x_0)$, then assuming $\pi_1(X, x_0)$ is abelian, for $[f]\in\pi_1(X, x_0)$, $$\begin{eqnarray} \widehat{\alpha}([f])&=& [\overline{\alpha}]*[f]*[\alpha]\\ &=& ([\overline{\beta}]*[\beta])*([\overline{\alpha}]*[f]*[\alpha])*([\overline{\beta}]*[\beta])\\ &=& [\overline{\beta}]*([\beta]*[\overline{\alpha}])*[f]*([\alpha]*[\overline{\beta}]*[\beta])\\ &=& [\overline{\beta}]*[f]*([\beta]*[\overline{\alpha}])*([\alpha]*[\overline{\beta}]*[\beta]) \;\;\;\;\{\textrm{because } [f], [\beta]*[\overline{\alpha}]\in\pi_1(X, x_0)\}\\ &=& [\overline{\beta}]*[f]*[\beta]\\ &=& \widehat{\beta}([f]). \end{eqnarray}$$

Prajwal Kansakar
  • 5,391