Question concerning the notation of path multiplication

Question

I'm self-studying topology by using Lee's Introduction to topological manifolds. I've just started reading the chapter on Homotopy and the Fundamental Group. Untill now everything makes perfect sense to me. The only thing bothering me is the notation for the path multiplication, which is \begin{equation} fg(s):=\begin{cases}f(2s), s\in [0,1/2] \\ g(2s-1), s \in [1/2,1]\end{cases} \end{equation} first going along $f$ and then $g$. Because composition of functions $f:X\to Y, g:Y \to Z$ has always been denoted by $g\circ f:X\to Z$ I'd find it much more compatible to denote the path multiplication by $gf$ instead of $fg.$

So my question is why this notation is used?

Answer 1

I think the simplest answer is that the notation for composition of functions is highly unnatural. We write compositions the way we do only because of the influence of the traditional $f(x)$ notation for applying a function to an argument. We all had to work to get used to the fact that functional composition works "backwards," with $f\circ g$ meaning "first apply $g$, then apply $f$." As several others have noted, various authors have tried to "correct" that problem by writing functional notation in the opposite order, as $xf$ or $(x)f$. But ingrained habits don't get overturned so easily.

But path multiplication is not composition of functions, so there's no reason to make the notation for path multiplication follow the same perverse pattern as functional composition. The notation $fg$ or $f\cdot g$ means "first follow $f$, then follow $g$," just as it appears.

Answer 2

One reason for the notation is the prevalence of graphical techniques used for gaining intuition, teaching or even as mathematical proof. For instance the following diagram represents the homotopy giving associativity for loop multiplication $(f\cdot g)\cdot h\simeq f\cdot (g\cdot h)$

In the diagram the left-most region represents the loop $f$, the middle region represents the loop $g$ and the right-hand region represents $h$. Clearly it displays from left to right and leads to adopting the notation $f\cdot g \cdot h$ for the loop $f$ followed by $g$ followed by $h$.

And if you're not happy with that answer then do what I would normally do and blame it all on poor old Henri Poincaré, http://www.maths.ed.ac.uk/~aar/papers/poincare2009.pdf page 58.

Answer 3

Just a couple of days ago, I revisited this exact question and thought "I'm just going to do path multiplication in the 'correct' order." What happened:

• The universal covering space of a space $X$ can be constructed using all homotopy classes of paths starting from the basepoint $x_0$ (rel endpoints), and the deck transformation group is $\pi_1(X,x_0)$. The action for $[\gamma]\in \pi_1(X,x_0)$ on a path $[\alpha]$ is $[\gamma][\alpha]=[\gamma\alpha]$, which would have to be $[\alpha\gamma^{-1}]$ in the other order to get a group homomorphism. (Though the opposite order works fine if deck transformations are a right action instead.)

• This one is hard to explain quickly: given a knot complement ($S^3-K$ for a knot $K$), the twisted 1-cocycles associated with to a map $\phi:\pi_1(S^3-K)\to\mathbb{C}^\times$ correspond to affine representations of $\pi_1(S^3-K)$. In particular, if $f:\pi_1(S^3-k)\to \mathbb{C}$ is a $1$-cocycle, then $$\rho(g)=\begin{pmatrix}\phi(g)&f(g)\\0&1\end{pmatrix}$$ is a representation (using projective coordinates). I could only get this (or simple variations of this) to be a representation if path composition was in the standard "backwards" order. The cocycle condition is insensitive to the multiplication convention, and is somehow forcing the usual one.

I am guessing it has something to do with the cogroup structure of $S^1$, but I don't know enough to say.

Answer 4

The question of which order to write composition is a common issue. The book Categories and Groupoids uses $(x)f$ for evaluation of a function leading to composition of functions $(x)(f\circ g)=((x)f)g)$. This fits better with the arrow notation for functions, and for composition of paths, and is liked by algebraists. It also works out better in higher dimensional considerations.

The book Topology and Groupoids (T&G) uses, as did the 1968 edition, the notion of a path in $X$ of length $r \geqslant 0$ as a map $a:[0,r] \to X$, thinking of a path as a journey. This leads to composition of paths in $X$ forming a category $P(X)$, so there is no need for the picture proof given above by tyrone. However you still have to discuss reparametrisation, i.e. that any path is equivalent to a path of length $1$ (T&G Fig 6.6). See also the book Introduction to Knot Theory, which uses "Moore paths", i.e. pairs $(f,r)$ where $r \geqslant 0$ and $f:[0,\infty) \to X$ is constant on $[r, \infty)$. See also Wikipedia on Moore path space.

The intuitive notion of a path as a journey also makes it more natural to work in the fundamental groupoid $\pi_1(X,A)$ on a set of base points, rather than to concentrate the attention on "return journeys". See the discussion at Math Overflow.

The idea of a cogroup mentioned above is best thought of in terms of unit interval objects in Abstract Homotopy Theory.