I was hoping someone could review my proof. Thanks in advance!

Problem: Let x$_0$ and x$_1$ be points of the path-connected space X. Show that if for every pair $\alpha$ and $\beta$ of paths from x$_0$ and x$_1$, we have $\hat{\alpha}$ = $\hat{\beta}$ then $\pi_1$(X,x$_0$) is abelian.

Note: $\hat{\alpha}$ is the isomorphism from $\pi_1$(X,x$_0$) to $\pi_1$(X,x$_1$) via the usual map using a path from x$_0$ to x$_1$.


suppose $\hat{\alpha}$ = $\hat{\beta}$ and that X is path connected. Now since X is path connected we have that all fundamental groups, $\pi_1$(X,x$_j$) are isomorphic, for any x$_j$ $\in$ X. Suppose $\pi_1$(X,x$_0$) is not abelian. Then there exists and $f$,$g$ $\in$ $\pi_1$(X,x$_0$) such that f and g do not commute. Then $g$ is a loop from x$_0$ to x$_0$ but can be written as follows:

Let $\phi$ be the path from x$_0$ to $w$ for some $w$ $\in$ image($g$), where the path $\phi$ follows the loop $g$ up to some point $w$ $\ne$ x$_0$. And let $\delta$ be the path from x$_0$ to $w$ using the remainder of the loop $g$. Note that such a $w$ $\ne$ x$_0$ exists since if $g$ is the constant loop onto x$_0$ then $g$ = e$_{x_0}$ and g must commute with f, a contradiction to the case we are in.

Then we have $g$ = $\phi$ $\cdot$ $\bar{\delta}$. Then using the equivalence of $\hat{\phi}$ and $\hat{\delta}$ we have: $\bar{\phi}$ $\cdot$ $f$ $\cdot$ $\phi$ = $\bar{\delta}$ $\cdot$ $f$ $\cdot$ $\delta$, which implies that we have $f$ $\cdot$ $\phi$ $\cdot$ $\bar{\delta}$ = $\phi$ $\cdot$ $\bar{\delta}$ $\cdot$ $f$, which in turn implies $f$ $\cdot$ $g$ = $g$ $\cdot$ $f$, a contradiction. Hence the fundamental group with base point x$_0$ must be abelian.

Note that we have the equivalence of $\hat{\phi}$ and $\hat{\delta}$ as both are derived from paths from x$_0$ to $w$, with both points in X, and hence by hypothesis the isomorphisms they induce between the groups $\pi_1$(X,x$_0$) and $\pi_1$(X,$w$) are equivalent, namely $\hat{\phi}$ and $\hat{\delta}$. (is this a correct interpreation of the hypothesis or are x$_0$ and x$_1$ fixed? If so I believe we can just set x$_1$ = w?)

  • $\begingroup$ The part that reaches a contradiction is not necessary for this problem. You can obtain the property by direct implication. $\endgroup$ – Kevin. S Mar 17 at 6:12

Yes, your proof works, although it is a bit convoluted. Your proof can be paraphrased a bit more cleanly as follows:

The assumption is $$\forall x_0\xrightarrow{f}x_0 \forall x_0\xrightarrow{\alpha,\beta} x_1: \alpha\circ f\circ\alpha^{-1} \sim \beta \circ f\circ \beta^{-1}$$ Composing with $\alpha^{-1}$ from the left and with $\beta$ from the right side, this means $f\circ(\alpha^{-1}\circ\beta)\sim(\alpha^{-1}\circ\beta)\circ f$. Since every homotopy class $[g]\in \pi_1(X,x_0)$ can be written as $\alpha^{-1}\circ\beta$ for some paths $\alpha$ and $\beta$ from $x_0$ to $x_1$ (just fix any $\alpha$ and chose $\beta:=\alpha\circ g$), this proves that $f\circ g$ is homotopic to $g\circ f$ for all $[f],[g]\in\pi_1(X,x_0)$.

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