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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.

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  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Feb 15 '18 at 4:27
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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.

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  • $\begingroup$ 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, $[f]$ ? $\endgroup$ – user249018 Feb 12 '18 at 23:25
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    $\begingroup$ @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. $\endgroup$ – Alex Ortiz Feb 12 '18 at 23:29
  • $\begingroup$ 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 ? $\endgroup$ – user249018 Feb 12 '18 at 23:52
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    $\begingroup$ The fundamental group of the twice punctured plane is the free product $Z * Z$, not the direct product. $\endgroup$ – Alex Provost Feb 13 '18 at 0:34
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    $\begingroup$ @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. $\endgroup$ – Alex Provost Feb 13 '18 at 1:56

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