# Is distributivity sufficient to define composition?

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?

• 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)
– Mary
Commented Jan 7, 2015 at 2:40
• So what you are saying is if this basis consists of elementary functions, then all functions can be defined elementarily, so this definition works?
– k_g
Commented Jan 7, 2015 at 3:01
• 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.
– Mary
Commented Jan 7, 2015 at 3:21
• I see. But then composition would have to be defined to be distributive on the basis.
– k_g
Commented Jan 7, 2015 at 5:18
• 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 $x\mapsto 2x^2$. Commented Jan 10, 2015 at 3:51

Your conjecture is wrong. Let $q\colon A\to A$ be any map and define $f\odot g=f\circ q\circ g$. Then $$(f\star g)\odot h=(f\star g)\circ (q\circ h)=f\circ(q\circ h)\star g\circ(q\circ h)=f\odot h\star g\odot h$$
• @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. Commented Jan 10, 2015 at 22:19