Visualizing that fundamental group is not abelian in general

algebraic-topologyfundamental-groupshomotopy-theory

It is well-known that

The fundamental group of the figure eight is not abelian. i.e. the two ways of composing $a$ and $b$ are not homotopic to each other.

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Q1: Why are $ab$ and $ba$ not homotopic? (Edit: this has been answered here very nicely)

Added:
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Q2: Are $ab^{−1}$ in (A) and $ab$ in (B) homotopic?

Q3: Why $ab$ in (B) or in above picture is homotopic to a simple loop i.e. a circle? In other words why we are allowed to deform $ca$ in the below to its adjacent image? aren't they welded in the base point?

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img src: user326210

This is my thoughts: $ab$ is a loop based on the black point which is homotopic to a circle. (For better imagination I consider disc with two hole instead of figure 8.) In the following picture, $ab$ and $ab^{-1}$ are the concatenated loops. Look at the figure (A). there I think $ab^{-1}=b^{-1}a$. i.e. I think $b^{-1}a$ loop is exactly same as $ab^{-1}$. But what is wrong here? why they are not homotopic?

Note: I am aware about algebraic reasons of why they are not abelian. i.e. free groups with two generators is not abelian.. I just want to explore that what exactly is happening here. (and correct my misunderstandings)

Best Answer

The loops $ab^{-1}$ and $ab$ are not homotopic.

enter image description here

Imagine the holes are instead pegs, like the left figure here: CAD image of two pegs and two holes Source:https://www.tinkercad.com/things/11tjAfAiQNw-two-pegs-two-holes

The loop $ab^{-1}$ is equivalent to an open loop $\mathsf{O}$ around both pegs. First, you wrap string around the pegs to make the shape of $ab^{-1}$; tie the start and end of the string into a knot at the base point. Then, notice you can simply nudge the string into an $\mathsf{O}$ shape without moving the base point knot or lifting up the string.

In contrast, the loop $ab$ is different. If you wrap the string around the pegs to make the shape of $ab$, you create a figure $\mathsf{8}$. There is no way to nudge the string into an $\mathsf{O}$ shape without moving the base point or lifting up the string over the pegs.

"Aren't they welded to the base point". Note that you are allowed to nudge any part of the string except the knot where the string starts and ends. The string is allowed to cross over itself and cross over the base point. If part of the string crosses over itself at the base point, you can still move that part; just don't move the base knot itself.


You can think about homotopies like this to help your intuition. When you make any loop out of string, try nudging the string without (a) moving the base point, or (b) lifting up the string over the pegs. The result is another homotopically equivalent loop, and all homotopically equivalent loops can be made in this way.

The pegs are obstacles. Wrapping a string around them creates a loop that you can't remove unless you lift that loop over the peg. In this way, just by recording which strings can be homotopically transformed into other strings, you can discover where the pegs are, even if the pegs are invisible. Thus this loop-wrapping approach (homotopy theory) uses strings within the space to reveal the invisible obstacles/holes outside the space.

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