Is this a sufficient condition for a binary operation on functions to be a pointwise operation?

Question

Let $S$ be an arbitrary set, and let $F$ be the set of all functions from $S$ into $S$. A binary operation $*$ from $F \times F \rightarrow F$ is said to be a pointwise binary operation if there is a binary operation $+$ from $S \times S \rightarrow S$ such that $(f * g)(x) = f(x) + g(x)$, for all $f,g$ in $F$ and all $x$ in $S$. In an answer to a previous question, I read that a necessary condition for a binary operation $*$ to be a pointwise binary operation is: for all $f_1,f_2,g_1,g_2$ in $F$ and all $x$ in $S$, if $f_1(x)=f_2(x)$ and $g_1(x)=g_2(x)$, then $(f_1 * g_1)(x) = (f_2 * g_2)(x)$. Is that condition also a sufficient condition?

Answer 1

It isn't.

I will show that your condition is sufficient in case of something that I will call parametrized operator. Let me define this thing first: It's a function from $S\times S\times S\rightarrow S$. Now you will call a function $*$ from $F\times F\rightarrow F$ pointwise ternary if there is a parametrized operator $+$ such that $(f*g)(x)=+(f(x),g(x),x)$ for every $f,g$ and $x$. I will show that (1) for given $*$ which satisfies your condition, there is a (unique) parametrized operator $+$ which satisfies $(f*g)(x)=+(f(x),g(x),x)$ and I will also show (2) that for any given parametrized operator $+$ the operator $*$ defined by $(f*g)(x)=+(f(x),g(x),x)$ satisfies your condition. This will show us that there is a one-to-one correspondence (injectiveness is proved by (1) and surjectiveness is proved by (2)) between operators $*$ which satisfy your condition and parametrized operators.

Proof for (1):

We have $*$ which satisfy your condition. We will build (define) $+$ one triple at a time. Take arbitrary $(a,b,x)\in S\times S\times S$. Take some functions $f$ and $g$ from $F$ for which we have $f(x)=a$ and $g(x)=b$. There will be some that satisfy this. Now define $+(a,b,x)$ to be equal to $(f*g)(x)$. To show that this definition is valid, we have to prove that any choice we made on the way of defining it didn't influence our definition. For that let's take two different choices we could make. We only made choice while choosing the functions $f$ and $g$. So let's say in the first choice we chose $f_1$ and $g_1$ and in the second we chose $f_2$ and $g_2$. By the conditions under we were choosing, we have that $f_1(x)=f_2(x)=a$ and $g_1(x)=g_2(x)=b$. Now your condition tells us that $(f_1*g_1)(x)=(f_2*g_2)(x)$, which is exactly what we took as the definition of $+(a,b,x)$. This means our definition was valid indeed. It is straightforward to check that this $+$ satisfies $(f*g)(x)=+(f(x),g(x),x)$; call this (#). It is immediate from the way we defined $+$. So far we have proven that such parametrized operator exists, we are only left to show that it is unique for a given $*$. Suppose there are two different such parametrized operators $+_1$ and $+_2$. Let's say they differ for the triple $(a,b,x)$. Let's say $f$ and $g$ are such functions that $f(x)=a$ and $g(x)=b$. We have that $+_1$ and $+_2$ differ at $(f(x),g(x),x)$, i.e. $+_1(f(x),g(x),x)\neq+_2(f(x),g(x),x)$. But by (#) we also know that $+_1(f(x),g(x),x)=(f*g)(x)=+_2(f(x),g(x),x)$. It's a contradiction, we got that such $+$ is unique.

Proof for (2):

We have $+$ and we have $*$ defined from it by $(f*g)(x)=+(f(x),g(x),x)$. Let's say the assumtions of your conditions are satisfied. Namely, $f_1(x)=f_2(x)$ and $g_1(x)=g_2(x)$ for some $x$ and some functions $f_1,f_2,g_1,g_2$. By the definition of $*$ we have $(f_1*g_1)(x)=+(f_1(x),g_1(x),x)=+(f_2(x),g_2(x),x)=(f_2*g_2)(x)$ which is the hypothesis of your condition. Meaning, your condition is satisfied by a $*$ defined this way.

If you want this parametrized operator bound to your operator $*$ to be a true binary operator, then it must not depend on the third argument $x$. In other words, for all $a,b,x$ and $y$ we must have that $+(a,b,x)=+(a,b,y)$. It easily translates to the following change in your condition: For every $f_1,f_2,g_1,g_2$ from $F$ and every $x,y$ from $S$ for which $f_1(x)=f_2(y)$ and $g_1(x)=g_2(y)$ we have that $(f_1*g_1)(x)=(f_2*g_2)(y)$.

Easy counter-example exists whenever there are more parametrized operators than binary operators. This already happends for $|S|=2$.