Note: I will concatenate paths from left to right.

If $X$ is a path-connected topological space, and $p, q\in X$ are distinct, then any path $\gamma$ between $p$ and $q$ induces an isomorphism between $\pi_1(X, p)$ and $\pi_1(X, q)$, where a loop $\alpha$ based at $p$ is send to the loop $\gamma^{-1}\alpha\gamma$, which is based at $q$. If $\delta$ is another path between $p$ and $q$, then $\gamma$ and $\delta$ would induce the same isomorphism iff $\gamma^{-1}\alpha\gamma$ is homotopic (as a loops with fixed base point $q$) to $\delta^{-1}\alpha\delta$ for all $\alpha$, since these would then be in the same equivalence class and thus equal in $\pi_1(X, q)$. This would only seem to hold if every loop from $p$ to $q$ and then back to $p$ in the reverse direction (i.e., all paths of the form $\gamma^{-1}\gamma$ are homotopic.

Now, this condition is enough to answer the question, but I was wondering if it was equivalent to some simpler condition. I guess it's equivalent to all paths from $p$ to $q$ being homotopic to each other (while fixing $p$ and $q$). Is there a name for this condition?

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    $\begingroup$ If and only if $\pi_1(X)$ is abelian. $\endgroup$ – orangeskid Oct 13 '14 at 16:41
  • $\begingroup$ ^Why is that equivalent to any two paths between two points being homotopic? $\endgroup$ – Nishant Oct 13 '14 at 16:42
  • $\begingroup$ Proving what @orangeskid just said is a classic exercise in algebraic topology... you should try to prove it! $\endgroup$ – Mike Earnest Oct 13 '14 at 16:43
  • $\begingroup$ Okay! I'll get to it. $\endgroup$ – Nishant Oct 13 '14 at 16:43
  • $\begingroup$ Wait, is my conclusion that any two paths between $p$ and $q$ are homotopic incorrect? Because $S^1$ has the abelian group $\mathbb Z$ as its fundamental group, but two paths between $p$ and $q$ don't seem to be homotopic if they go around the circle a different number of times. $\endgroup$ – Nishant Oct 13 '14 at 16:51

It is equivalent to a simpler condition, but not quite the one you're thinking. Suppose you have two paths $\gamma_1, \gamma_2$ from $p$ to $q$. They induce isomorphisms $\phi_1, \phi_2$. Then the automorphism of $\pi_1(X,p)$ $\phi_2 \circ \phi_1^{-1}$ is just the inner automorphism given by conjugation by the loop $\gamma_1 \gamma_2^{-1}$. Note that any element of $\pi_1(X,p)$ can be realized as such a pair of loops (I leave this tothe reader). So for all isomorphisms to be the same, all inner automorphism of $\pi_1(X,p)$ must be trivial, which is precisely when $\pi_1(X,p)$ is abelian.

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One might as well state the algebraic result (as in Topology and Groupoids):

6.3.2 Let $x, x ′$ belong to the same component of the groupoid $G$. Then $a_*=b_*: G(x) \to G(x ′ )$ for all $a, b : x → x ′$ if and only if $G(x)$ is abelian.

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    $\begingroup$ @JHance: I've long accepted a maxim of Philip Hall which I learned from Philip Higgins that one should look for the algebra which models the geometry, and not try to force the geometry unnecessarily into an already given framework. I argue that 1-dimensional homotopy theory is better modelled by groupoids rather than by groups. $\endgroup$ – Ronnie Brown Oct 13 '14 at 21:11

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