Group homomorphism from $F(R)$ to $R$ which sends $f$ to $f(1)$. I am trying to prove the following:
Exercise:
Let $a$ be a function from $F(R)$ to $R$, defined by
$a(f)=f(1)$.
Note: I am assuming that $F(R)$ is the group of functions from $R$ to $R$.
Prove that $a$ is a homomorphism.
My thoughts so far:
I am not sure what are the operations on $F(R)$ and $R$, since they are not explicitly written.
Naturally I'd think that $F(R)$ would use composition and $R$ would use addition.
So now the following must be proven: $a(f \circ g)=a(f)+a(g)$.
Which is equivalent to $(1)+g(1)=(f \circ g)(1)$
But this is not true for all $f$ and $g$.
For example if we set $f$ and $g$ to be the identity we get:
$1+1=1$ which is false.
Another problem is that $F(R)$ is not abelian why $R$ is, so if it was a homomorphism,
then $(f \circ g)(1)=(g \circ f)(1)$, but that's obviously not true for all $f$ and $g$.
So if we look at $F(R)$ as a group of functions from $R$ to $R$ with addition it works:
(and also now we don't have to consider only bijections from $R$ to $R$)
So now we have:
$a(f + g)=a(f)+a(g)$
$(f+g)(1)=f(1)+g(1)$ which proves that it's a homomorphism.
So my question is:
Is it possible that the operation on $F(R)$ is not addition but something else?
And is it possible for a homomorphism to exist between a non-abelian and an abelian group?
 A: The set of functions from $\mathbb{R}$ to $\mathbb{R}$ has both an additive abelian group structure given by pointwise addition (where $(f+g)$ is the function whose value at $x$ is $f(x)+g(x)$), and also a ring structure also given by pointwise operations, (where $fg$ is the function whose value at $x$ is $f(x)g(x)$).
More generally, if $A$ is a group with operation $\odot$ (resp. ring with operations $+$ and $\cdot$), and $X$ is a set, then the collection of all functions from $X$ to $A$, denoted $A^X$, is an abelian group with pointwise addition, and a ring with pointwise operations. Indeed, we have that if $f$ and $g$ are functions from $X$ to $A$, then so are $(f\odot g)$ (resp. $f+g$ and $f\cdot g$). The operations are associative,
$$\begin{align*}((f\odot g)\odot h)(x) &= (f\odot g)(x)\odot h(x)\\ 
&= (f(x)\odot g(x))\odot h(x)\\
&= f(x)\odot(g(x)\odot h(x))\\
&=f(x)\odot ((g\odot h)(x))\\
&= (f\odot(g\odot h))(x).
\end{align*}$$
Since $f\odot(g\odot h)$ and $(f\odot g)\odot h$ take the same values at every $x\in X$, they are equal as functions.
Similarly with $+$ and $\cdot$. The function that functions as an identity for $\odot$ is the function that sends every $x\in X$ to the $\odot$-identity of $A$ (to $0$ in the ring); the “inverse” of the function $f\colon X\to A$ is the function that sends $x\in X$ to $(f(x))^{-1}$ in $A$.
Now fix $x_0\in X$. Verify that in this group/ring structure, the map that sends the function $f$ to the element $f(x_0)$ is a group/ring homomorphism.
What you have is a special case opf this general construction.
(Even more abstractly, $A^X$ is nothing more than $\prod_{x\in X}A$, and the map $f\mapsto f(x_0)$ is the production onto the $x_0$th coordinate; viewed this way it is clear that if $A$ is any algebra, in the sense of universal algebra, then the set of functions $X\to A$, for an arbitrary set $X$, is also an algebra of the same type, satisfying the same set of identities as $A$ does, and that the map sending each function $f$ to $f(x_0)$ for a fixed $x_0\in X$ is a homomorphism from $A^X$ to $A$)
