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Simple question but what is the definition that allows me to take (assume everything that is also needed for this proof is here) $(f \circ g) \circ h(w)$ and turn it into $f(g(h(w)))?$

I see this used a lot in function proofs, but I'm not sure exactly which definition it comes from.

George
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  • https://math.stackexchange.com/a/523929/516898 – 1Spectre1 Dec 12 '18 at 08:59
  • Just carefully unwind the definition of function composition twice. $f \circ g$ is, by definition, the function with the property that $(f \circ g)(x) = f(g(x))$. Similarly, $(f \circ g) \circ h$ is, by definition, the function with the property that $((f \circ g) \circ h)(w) = (f \circ g)(h(w)) = f(g(h(w))$. – Qiaochu Yuan Dec 12 '18 at 09:12

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Associativity follows almost immediately from the definition of composition of functions.

Since $(f \circ g)(x) = f(g(x))$ we have

$((f \circ g) \circ h)(w) = (f \circ g) (h(w))= f(g(h(w))) \space \forall w \in W$

where $W$ is the domain of $h$, and

$(f \circ (g \circ h))(w) = (f(g \circ h))(w) = f(g(h(w)))\space \forall w \in W$

Since $((f \circ g) \circ h)(w) = ((f \circ (g \circ h))(w) \space \forall w \in W$ we can conclude that $(f \circ g) \circ h = f \circ (g \circ h)$

gandalf61
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