Do functions with composition form a ring?

Question

Functions with composition form a monoid, and injective functions with composition form a group. Is there in some sense a natural or useful definition for "addition" that would enable formation of a ring? If not, is there a sensible subset of functions where this is possible?

Answer 1

If $X$ is a set with more than one element and $M$ is the set of functions $X\to X$, then there is no addition operation on $M$ that makes $M$ a ring where composition is multiplication. Indeed, if there were such an operation, then the $0$ element of $M$ would need to satisfy $0f=f0=0$ for all $f\in M$. But there is no such function $0$. For instance, pick some $x\in X$ and pick some $y\in X$ which is different from $0(x)$ (we can do this since $X$ has more than one element). Let $f$ be the constant function with value $y$. Then $f0(x)=y\neq 0(x)$ so $f0\neq 0$.

One variant where you can get a nice ring structure is when $X$ itself has an abelian group structure, and you restrict $M$ to consist of only the group homomorphisms $X\to X$. Then you can define an addition on $M$ by pointwise addition ($(f+g)(x)=f(x)+g(x)$ where on the right side we use the group operation of $X$)), and this will make $M$ a ring with composition as multiplication, called the endomorphism ring of $X$.