# Isomorphism equivalence implies fundamental group is abelian

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$$_1$$.

Solution:

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?)

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

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 $$[f],[g]\in\pi_1(X,x_0)$$.