How to read $(h \circ g)\circ f = h \circ (g \circ f)$?

Question

Let $f, g, h$ be functions. I learned that $g \circ f$ is read by "the composition of $f$ and $g$".

Then, how to read $(h \circ g)\circ f = h \circ (g \circ f)$?

the composition of $f$ and 'the composition of $g$ and $h$' is equal to the composition of 'the composition of $f$ and $g$' and $h$?

I thought it is precise, but too long and not intuitive, and also hard to understand.

Answer 1

Associativity is a bit of a funny one, because it's a statement that "the operation is simple to understand". Statements along the lines of "something is simple" are often weirdly confusing, because not only do they encompass the possibility of the something being confusing, but also because there isn't a world in which anything is actually confusing; so you have no concrete frame of reference in which to understand the negation of the statement.

The left-hand side, $(h \circ g) \circ f$, means "do $f$, then do the composition $h \circ g$"; that is, "do $f$, then do (do $g$, then do $h$)". The right-hand side is "do (do $f$, then do $g$), then do $h$". Of course, both of those are really "do $f$ then $g$ then $h$".

---

If by "how do I read" you mean "how do I pronounce", then I'd actually use hand signals with this, by waving brackets using my hands.

[form pair of brackets with hands] $h$ after $g$ [slight pause, stop forming brackets with hands] after $f$ is equal to $h$ after [slight pause, form brackets, speak faster] $g$ after $f$ [stop forming brackets].

Or since there's a strong convention around this, you can simplify the pronunciation of "$g \circ f$" (and so on) by pronouncing it "$gf$" and just not mentioning the composition operator at all; after all, what other operator would it be?

Answer 2

If you start with $3$ functions $f:A\to B$, $g:B\to C$ and $h:C\to D$ then you can construct two functions $A\to D$.

We start with constructing $u:=h\circ g:B\to D$ and $v:=g\circ f:A\to C$.

The second step is constructing $u\circ f:A\to D$ and $h\circ v:A\to D$.

In this context it is said that the constructed functions $u\circ f$ and $h\circ v$ will be the same.

This reveals itself by unwrapping the prescription according to the general rule: $$p\circ q(x):=p(q(x))$$

Observe that for every $a\in A$: $$[u\circ f](a)=u(f(a))=h(g(f(a)))=h(v(a))=[h\circ v](a)$$