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I am having difficulty in distinguishing between two equivalence classes of the fundamental group at a base pont $x_0$ of a topological space $X$. Given $X$ and an arbitrary point $x_0\in X$ one defines homotopy as equivalence relation on the set of functions on $X$ with base point $x_0$ called loops. In other words, all loops at a fixed base point $x_0$ seem to me to be homotopic, and thus belonging to one equivalence class, say [f], since two such loops seem to be continuously deformable into one another.

Can somebody help me understand to discriminate between two equivalence classes, [f] and [g], modulo the homotopy of loops at a base point $x_0$ ?

Many thanks.

user249018
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  • Why do you think that "two such loops seem to be continuously deformable into one another"? Either you are misunderstanding some definition or aren't thinking about the right examples, but it's very hard to tell what you're missing since you haven't explained any of your beliefs. – Eric Wofsey Feb 15 '18 at 04:27

1 Answers1

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Imagine the punctured plane $\Bbb R^2\setminus \{0\}$. Any loop that encircles the origin is not homotopic to a loop that doesn't contain the origin, though they may share the same base point. Intuitively, any loop that doesn't contain the origin can be shrunk to a point because there is no hole to obstruct it during its homotopy, but loops that do encircle the origin can't be shrunk to a point because we would have to "tear" them at some point in the homotopy.

This isn't precise, but you're studying algebraic topology to learn how to make it precise. More precisely, we would say that $\pi_1(\Bbb R^2\setminus\{0\},(1,0))\cong \Bbb Z$ is a nontrivial group that is generated by an element $[f]$ (such as a loop that encircles the missing origin). Any loop $g$ that does not encircle the origin is nulhomotopic, so its homotopy class $[g]$ is trivial, so we can't have $[f]=[g]$ because $\Bbb Z$ is a nontrivial group.

Alex Ortiz
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  • Thanks. Your example clarified many things. So, in the case you are talking about, $[f]\neq [g]$, but $[g] \in \pi_1(\Bbb R^2\setminus{0},(1,0))$, right ? In the case there is no point missing of $\Bbb R^2$, can one say that $\pi_1$ has only one element, $[f]$ ? – user249018 Feb 12 '18 at 23:25
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    @user249018 Yes, $[g]\in\pi_1(\Bbb R^2\setminus{0},(1,0))$. It's true that $\pi_1(\Bbb R^2) = 0$, the trivial group, and hence has only one element. This is just the statement that $\Bbb R^2$ is simply connected. – Alex Ortiz Feb 12 '18 at 23:29
  • To add to my understanding, in the case there are two different points of $\Bbb R^2 $ missing, say $x $ and $y$, can we then say that there are in total $4$ equivalence classes, i.e. $\pi_1(\Bbb R^2\setminus{x, y},(1,0))$ has $4$ elements as homotopy classes ? Can you maybe give the value of $\pi_1$ in this case ? – user249018 Feb 12 '18 at 23:52
  • @user249018 It's not quite so obvious what $\pi_1(\Bbb R^2\setminus{x,y})$ is. It turns out that $\Bbb R^2\setminus{x,y}$ is homotopy equivalent to the "figure-eight" space ($8$), whose fundamental group is actually $\Bbb Z\times\Bbb Z$ (i.e. two generators) – Alex Ortiz Feb 12 '18 at 23:59
  • But are there $4$ homotopy classes in $\pi_1(\Bbb R^2\setminus{x,y}) $, since I think there are 4 different loops which cant be continuously deformed into each other ? When you say that $\pi_1(\Bbb R^2\setminus{0},(1,0)) $ is generated by an element $[f], $ how does this generator then generate the nulhomotopic element $[g]$ ? By saying generator I have in mind group theoretical prescription though. I will highly appreciate your answers. – user249018 Feb 13 '18 at 00:07
  • I am remaining perplexed in thinking of 4 equivalence classes of $\pi_1(\Bbb R^2\setminus{x,y}) $. Do you at least agree that there are 4 sets of loops, which cant be continuously deformed into each other ? – user249018 Feb 13 '18 at 00:23
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    The fundamental group of the twice punctured plane is the free product $Z * Z$, not the direct product. – Alex Provost Feb 13 '18 at 00:34
  • @Alex Provost. Can you maybe explain if there are 4 homotopy classes or not in the twice punctured plane ? Thanks. – user249018 Feb 13 '18 at 00:50
  • @user249018 You might be thinking of two generators (loops around the punctures) along with their inverses. But there are many other classes in the fundamental group, obtained by concatenating these loops in any arbitrary manner. – Alex Provost Feb 13 '18 at 01:10
  • @Alex Provost. Is the set of loops encircling the two punctures at the same time a homotopy class in itself ? I was thinking of this class and three others remaining when I said there are 4 of them. Can one not concatenate in the once punctured plane, because as I understood, in $\pi_1(\Bbb R^2\setminus{0},(1,0))$ there were only two equivalence classes, $[f], [g], $ of which $[g]$ was the nullhomotopic ? – user249018 Feb 13 '18 at 01:28
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    @user249018 I'm not sure I understand your first question. If you concatenate a loop encircling both punctures with itself, you obtain a different (non-homotopic) loop. And no, the fundamental group of the once punctured plane is infinite cyclic; looping $n$ times around the circle in a positive orientation corresponds to $n$ times the positive generator. – Alex Provost Feb 13 '18 at 01:56
  • To put it short, let $[s]$ be the set of loops encircling the two punctures $x$ and $y$ from the same base point $x_0.$ Is $[s]\in \pi_1(\Bbb R^2\setminus{x,y}) $ ? – user249018 Feb 13 '18 at 02:32
  • @user249018 That doesn't make sense. The notation $[s]$ stands for the homotopy class of $s$, and you can't have two non-homotopic loops in the same class. So $[s]$ cannot equal the set of all loops encircling the two punctures. – Alex Provost Feb 13 '18 at 02:56
  • @user249018 For instance, consider a loop $s$ going once around both punctures. This represents a non-trivial class $[s] \in \pi_1(R^2 - {x,y})$. But the concatenation $[s \cdot s]$ represents a different element in this group (it goes around the punctures twice), i.e. $[s] \neq [s \cdot s]$ even though both are loops that encircle both punctures. – Alex Provost Feb 13 '18 at 02:59
  • @Alex Provost Thanks. So $[s]$ is the homotopy class of loops encircling once the two punctures, $[s^n]$ is then the homotopy class of loops encircling n-times the two punctures, and both $[s], [s^n]$ are elements of $ \pi_1(R^2 - {x,y}).$ I hope this is now correct. The very last thing: Whats the generator of $[s]$ and $[s^2]$ if $[f]$ and $[k]$ are both generators of $ \pi_1(R^2 - {x})$ and $ \pi_1(R^2 - {y})$, respectively ? – user249018 Feb 13 '18 at 03:51
  • @user249018 Yeah, the first bit is correct. If you imagine the basepoint lying between $x$ and $y$, you can think of $s$ as the concatenation of a loop around the first puncture with a loop around the second puncture, i.e. $[s] = [f][k]$ in your notation. More generally, the statement "the fundamental group of the twice punctured plane is the free product of two infinite cyclic groups" means that any homotopy class can be represented as a product of $[f], [k], [f]^{-1}, [k]^{-1}$. The order is important : $[f][k]$ is not the same as $[k][f]$ (otherwise the group would be $Z \times Z$). – Alex Provost Feb 14 '18 at 13:21
  • I'm studying the fundamental group of the circle, which is said to be homeomorphic to $\mathbb{Z}$. If m and n are distinct integers, what would make a loop going around the circle n times different from one going around the circle m times? – ensbana Dec 06 '19 at 13:09