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Suppose we are in a path connected space $X$ , then the homotopy classes of loops about any point has a bijection between. Even if one takes the whole space and quotients the loop by free homotopy(*), then the resulting set of homotopy classes of free homotopy would be in bijection with the homotopy class of the loop about any point.

This makes me wonder then if we can define a composition on them without so we can get a fundamental group of the space $X$ without reference to any particular point.

My try:

Consider the set of loops on the space w.r.t free homotopy. Define the composition $*$ of two loops $\sigma_1 : S^1 \to X$ and $\sigma_2: S^1 \to X$ as follows. We pick a point $x_1 \in \sigma_1 (S^1)$ and a point $x_2 \in \sigma_2(S^1)$. I believe then we define we pick a point on each curve, use path connectedness to make a path between em, and then define a new curve where one traverses $\sigma_2$, $\gamma$ and $\sigma_1$ and say it's continous through application of pasting lemma.

However, I am not exactly sure how to define the composition loop such that everything is well defined exactly though.

How do I complete this approach? Is it even a fruitful approach?

*:Two loops $\sigma_1: S^1 \to X \sim \sigma_2 :S^1\to X$ are free homotopic when there exist a function $H: S^1 \times [0,1] \to X$ with $H( \_,0) = \sigma_1(\_)$ , $H(\_,1) = \sigma_2( \_ )$

Clemens Bartholdy
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  • The commutator of two free loops is freely null-homotopic. Hence your group will end up being abelian. This will end up being the first homology group of the space, which is the abelianization of the fundamental group. – Cheerful Parsnip Sep 21 '24 at 15:25
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    There is a notion of fundamental groupoid where you work with paths and there is no distinguished basepoint. – Moishe Kohan Sep 21 '24 at 15:54

1 Answers1

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The set of free homotopy classes of loops in a path-connected space $X$ can be identified, canonically, with the set of conjugacy classes of the fundamental group $\pi_1(X, x)$, at any basepoint. There is no way to define a natural multiplication on this set; if you try to make your idea work I believe you'll find that the result depends on the choice of path you use to connect the two loops.

However, you can instead use the fundamental groupoid $\Pi_1(X)$. This is the groupoid whose objects are given by all the points of $X$ and whose morphisms are given by homotopy classes of paths between points. Up to equivalence it contains the same information as the fundamental group, but because it does not depend on a choice of basepoint you can do things with it that you can't with the fundamental group.

For example, if $G$ is a group acting on $X$ then it induces, functorially, an action on $\Pi_1(X)$. This is not true for the fundamental group: to get an action on the fundamental group would require that $G$ fixes a point, and it is possible for $G$ to not fix any point, even up to homotopy in a suitable sense. In fact there are actions of $G$ on $\Pi_1(X)$ that are not equivalent to actions of $G$ on $\pi_1(X, x)$. I describe this in more detail (with an example) here.

Qiaochu Yuan
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