Let $A$ be a ring and $S$ be a set. Let $F(S,A)$ be the set of functions of the form $f: S \rightarrow A$. For all $f,g \in F(S,A)$ define $(f+g)(s)=f(s)+g(s)$, and $(fg)(s)=f(s)g(s)$.
Show that $F(S,A)$ is commutative if and only if $A$ is commutative.
If $A$ is commutative then for all $f,g\in F(S,A)$ and all $s\in S$ we have $$(fg)(s)=f(s)g(s)=g(s)f(s)=(gf)(s),$$ which means that $fg=gf$, and hence $F(S,A)$ is commutative.
If $A$ is not commutative there exist $x,y\in A$ such that $xy\neq yx$. For the maps $c_x$ and $c_y$ mapping constantly to $x$ resp. $y$ it holds $c_xc_y\neq c_yc_x$, so $F(S,A)$ is not commutative if $S\neq\varnothing$.
EDIT (thanks to Paolo Capriotti):
If $S=\varnothing$ then $F(S,A)=\{\varnothing\}$ which is commutative, so the statement is false.