I know $\pi_1(\mathbb{R}^2\setminus\{x,y\}) = \mathbb{Z}\ast\mathbb{Z} = \langle a,b\rangle$, but its non-abelian-ness isn't obvious to me.
Specifically, I draw a box and two points to represent $x$ and $y$. I call the clockwise loop around $x$ $a$, and the clockwise loop around $y$, $b$.
If I draw the loops $ab$ and $ba$ they look `obviously homotopic' to me (they are both a single clockwise loop containing both $x$ and $y$), so should be the same element in $\pi_1$, which is clearly not the case if we take the free product of $\mathbb{Z}$ with itself.
It seems that I must be making a big mistake somewhere. What is the obstruction to $ab \simeq ba$? I would prefer something concrete, I know this fundamental group follows from standard theorems (van Kampen's theorem), but I want to see the homotopy fail, because I think they are pretty clearly homotopic – so there is something wrong with my intuition which I'd like to fix.
Best Answer
An important detail is that the fundamental group is built from loops that all start and end at a common base point. We know that some loops can be continuously deformed into other loops; these loops are called homotopically equivalent. As a special case, some loops can be continuously deformed into other loops while keeping the base point fixed in place.
The fundamental group consists of loops which are considered equivalent when they are homotopic in this special base-point-preserving sense. The fundamental group fails to be commutative when there are loops that are homotopically equivalent in the generic sense, and yet not homotopically equivalent in the base-preserving sense.
In pictures, below you can see the plane with two holes along with three loops a, b, c with a common base point.
Note that you can't continuously deform b into c unless you're allowed to move the base point. (This is our signal that the fundamental group doesn't commute, which we'll show in more detail.)
Even so, b and c are homotopic in the general sense. For example, one (non-base-point-preserving) homotopy from b to c involves sliding the base point around the lower hole:
A neat observation is that during this homotopy, the trajectory of the base point is recognizable as loop a!
From this, it follows that there's a base-preserving homotopy between loop c•a ("go around the hole, then perform c") and loop a•b ("perform b, then go around the hole"):
If the fundamental group were abelian, then this conjugacy would imply that b = c. Because b ≢ c, the fundamental group is nonabelian.
In general, when there's a homotopy between two loops b and c, let a be the trajectory of the base point under that homotopy. Then a is itself a loop with the same base point as b and c, and c= aba⁻¹ (b and c are conjugates). If there isn't a base-preserving homotopy, then a is nontrivial and b ≢ c, so the fundamental group isn't abelian.
Final comment: Note that from the beginning, the source of the problem (noncommutativity) is the presence of holes. We saw intuitively that we could deform b into c, but only by moving the base point— there's a hole in the way. Now we can see that property emerge algebraically:
Loops b and c are deformable into one another when they're conjugate c= aba⁻¹. The element a describes the trajectory of the base point during that deformation. So if they're deformable without moving the base point, this is equivalent to saying that conjugacy holds when a is trivial (topologically, a is the constant loop at the base point; algebraically, it's the multiplicative unit) in which case we have b = c. Otherwise, if we must move the base point in order to deform one loop into the other, this is equivalent to saying that conjugacy c= aba⁻¹ only holds for a nontrivial choice of loop a. A loop is nontrivial just when it encircles a hole (!), hence this is an algebraic way of saying that a hole (call it hole A) stands in the way of deforming b into c.