Function Composition has the property of distributivity:

$$(f\star g)\circ h = (f\circ h)\star(g\circ h)\;\forall f,g,\star \in\{+,-,\times,\div\}$$

I was wondering if these properties uniquely define composition.

Intuitively, this makes sense. For example:

$$(x\mapsto x^2)\circ f = (I\times I)\circ f = I\circ f \times I\circ f = f^2$$

and a similar process could be defined for any function.

But does this work when functions cannot be easily defined in terms of elementary operations?

  • $\begingroup$ To compute the composition easily as suggested you would benefit from an Hamel basis for the space of functions en.m.wikipedia.org/wiki/Schauder_basis (see related concepts section) $\endgroup$
    – Mary
    Commented Jan 7, 2015 at 2:40
  • $\begingroup$ So what you are saying is if this basis consists of elementary functions, then all functions can be defined elementarily, so this definition works? $\endgroup$
    – k_g
    Commented Jan 7, 2015 at 3:01
  • $\begingroup$ If you know the Hamel basis, then you can express any function as a finite linear combination of functions and this would simplify the computation. Otherwise you can try to use a Taylor series, but this could yield infinite countable calculations. $\endgroup$
    – Mary
    Commented Jan 7, 2015 at 3:21
  • $\begingroup$ I see. But then composition would have to be defined to be distributive on the basis. $\endgroup$
    – k_g
    Commented Jan 7, 2015 at 5:18
  • 1
    $\begingroup$ Left distributivity (your first condition) is certainly a property of composition, but right distributivity is most certainly not! $$(x\mapsto x^2)\circ ((x\mapsto x)+(x\mapsto x)) \neq ((x\mapsto x^2)\circ (x\mapsto x))+((x\mapsto x^2)\circ (x\mapsto x)). $$ The left side is $x\mapsto 4x^2$. The right side is $x\mapsto 2x^2$. $\endgroup$ Commented Jan 10, 2015 at 3:51

1 Answer 1


Your conjecture is wrong. Let $q\colon A\to A$ be any map and define $f\odot g=f\circ q\circ g$. Then $$ (f\star g)\odot h=(f\star g)\circ (q\circ h)=f\circ(q\circ h)\star g\circ(q\circ h)=f\odot h\star g\odot h$$

  • $\begingroup$ OK. Thanks. Could adding another property uniquely define composition? $\endgroup$
    – k_g
    Commented Jan 10, 2015 at 20:23
  • $\begingroup$ @k_g Of course $\circ$ is associative and my $\odot $ in general is not. But I am unsure if demanding associativity would be enough - composition is so very specific. $\endgroup$ Commented Jan 10, 2015 at 22:19

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