My intent is to show that a composition of bijections is also a bijection by showing the existence of an inverse. But my approach requires the associativity of function composition.
Let $f: X \rightarrow Y, g: Y \rightarrow Z, h: Z \rightarrow W$ be functions.
$((f \circ g) \circ h)(x) = h((f \circ g)(x)) = h(g(f(x)))$, and $(f \circ (g \circ h))(x) = (g \circ h)(f(x)) = h(g(f(x)))$.
However, I am having problems in justifying that the two compositions, $(f \circ g) \circ h$ and $f \circ (g \circ h)$, have the same domain and range. When I consulted ProofWiki, whose link is at the bottom, I got even more confused. Specifically, for $(f \circ g) \circ h = f \circ (g \circ h)$ to be defined, ProofWiki requires that dom$g =$ codom$f$ and dom$h =$ codom$g$.
First of all, I think that it should be dom$g =$ range$f$ …. Moreover, as you can see in the example below, you actually have to adjust domains and ranges of $f, g, h$ for the requirement to hold true.
Let $f: \mathbb R \rightarrow \mathbb R$ be $f(x) = 2x$, $g: \mathbb R^+ \rightarrow \mathbb R$ be $g(y) = ln(y)$, $h: \mathbb R \rightarrow \mathbb R$ be $h(z) = z – 10$.
Then $((f \circ g) \circ h)(x) = ln(2x) – 10 = (f \circ (g \circ h))(x)$, with dom$((f \circ g) \circ h) = \mathbb R^+$ = dom$(f \circ (g \circ h))$. As a result, we need to set dom$f = \mathbb R^+$, range$f = \mathbb R^+$; dom$g$, range$g$, dom$h$, and range$h$ remain the same. Am I allowed to do that?
This adjustment implies that when we say dom$f = X$, $f$ must be defined for all elements in $X$, but $X$ may not be the entire set of elements for which $f$ is defined.
http://www.proofwiki.org/wiki/Composition_of_Mappings_is_Associative
Best Answer
Usually, when $f\colon X\to Y$ and $g\colon Y\to Z$ are maps, their composition is written $g\circ f$, rather than $f\circ g$: in this way you write $$ g\circ f(x)=g(f(x)) $$ by definition.
You seem to confuse codomain and range. The range, or image, of $f$ is the subset of the codomain $Y$ consisting of the elements $f(x)$, for $x\in X$. The range has no role whatsoever when composition of maps is considered. At least, when maps are supposed to be defined on the whole domain as is the case when talking of surjectivity or bijectivity.
Associativity is almost obvious. If you have another function $h\colon Z\to W$, you have, by definition, that $g\circ f\colon X\to Z$ and $h\circ g\colon Y\to W$. Thus one can consider also the compositions $$ h\circ(g\circ f) \qquad\text{and}\qquad (h\circ g)\circ f $$ and both are maps $X\to W$, so it makes sense to ask if they are equal. They are, because for each $x\in X$ we have $$ h\circ(g\circ f)(x)=h(g\circ f(x))= h(g(f(x))=h\circ g(f(x))=(h\circ g)\circ f(x). $$ If you can't parse this, just set $y=f(x)$, $z=g(y)$, $F=g\circ f$ and $G=h\circ g$, so that $F(x)=g(f(x))=g(y)=z$. Then $$ h\circ(g\circ f)(x)=h\circ F(x)=h(F(x))=h(z) $$ and $$ (h\circ g)\circ f(x)=G\circ f(x)=G(y)=h\circ g(y)=h(g(y))=h(z) $$ so the two elements are the same.