Homotopy/fundamental group question: Why group axioms fail when defined on paths?

Question

Neither Munkres nor Lee in their textbooks explicitly show why (fundamental) group properties like associativity fail when defined at the level of paths but work fine for homotopy classes of paths.

Given three paths $f$, $g$, and $h$, and if they are composable, why is the following ill-defined: $f.(g.h) = (f. g).h$.

Why should we only consider $[f], [g], [h]$? What exactly goes wrong with defining it at the level of paths?

Thank you.

Answer 1

The two functions $(a\cdot b)\cdot c$ and $a\cdot(b\cdot c)$ are;

$$\begin{align}((a\cdot b)\cdot c)(t)&=\begin{cases}a(4t)&t\in[0,1/4]\\ b(4t-1)&t\in[1/4,1/2]\\ c(2t-1)&t\in[1/2,1] \end{cases}\\ (a\cdot (b\cdot c))(t)&=\begin{cases}a(2t)&t\in[0,1/2]\\ b(4t-2)&t\in[1/2,3/4]\\ c(4t-3)&t\in[1/4,1] \end{cases} \end{align}$$

These two paths are rarely exactly the same.

Answer 2

Your original question was answered. The purpose of this answer is to provide a broader context to the question, and perhaps in some sense answer it differently. The problem is not so much that the group properties fail, but that taking paths instead of homotopy classes of paths is a different object, measuring different aspects of the space. Firstly, simply taking all loops $S^1\to X$ for a pointed space $X$ and taking as binary operation $f\circ g$ to be some concatenation of the two paths fails to be a group, but it is a very reasonable structure. It fails to be a group since you must choose a particular reparametrisation to obtain the concatenation, and that will fail to be associative. It will however be associative up to higher constraints which themselves will satisfy higher associativity constraints in an infinite, and quite intricate, tower. So it makes perfect sense to take the paths themselves, but you get something more complicated.

There is a very simple way to avoid some of these difficulties by considering loops to be maps $[0,a]\to X$ for all possible intervals $[0,a]$. Then a canonical concatenation of $[0,a]\to X$ and $[0,b]\to X$ is naturally given by the obvious identification of the pushout of the two intervals as $[0,a+b]$. This operation is associative and unital (the unit is the constant path $[0,0]\to X$) but lacks inverses (though there is a canonical choice for a homotopy inverse). So this is a monoid, and it holds much more information than the fundamental group. But that is precisely the problem. It's a much more complicated and larger object, so more difficult to compute. The quest is not necessarily to find a group but rather to consider the idea of homotopy, the information lost when passing to homotopy, and see what you get. You get a group, and that makes us very happy. There are tools to compute that group, and that makes us even a happier.