Here's the problem I did for homework from A First Course in Abstract Algebra, 7th Edition by John B. Fraleigh. I just want to check if my reasoning is correct on problem number 15 from Section 3:
Let $F$ be the set of all functions $f$ mapping $\mathbb{R}$ into $\mathbb{R}$ that have derivatives of all orders. Determine whether the given map $\phi$ is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not?
$\langle F,+\rangle$ with $\langle F,+\rangle$ where $\phi(f)(x)=x\cdot f(x)$
Given the binary structures $\langle S,*\rangle$ and $\langle S',*'\rangle$, in our class we define an isomorphism $\phi:S\to S'$ as a one-to-one correspondence such that $\phi(x*y)=\phi(x)*'\phi(y)$ for all $x,y\in S$.
I checked whether $\phi$ in the given problem is injective and concluded it is not by doing the following:
Assume $\phi(f)(x)=\phi(g)(x),\exists f,g\in F.$
$\begin{align} x\cdot f(x)&=x\cdot g(x)\\
x\cdot f(x)-x\cdot g(x)&=0 \\
x(f(x)-g(x))&=0\end{align} $
$\implies x=0$ or $f(x)=g(x)$. Since $f(x)=g(x)$ need not hold if $x=0,$ $\phi$ is not injective. Therefore, $\phi$ is not an isomorphism since it is not a one-to-one correspondence.
However, the solution manual said:
It is not an isomorphism because $\phi$ does not map $F$ onto $F$. Note that $\phi(f)(0)=0\cdot f(0)=0.$ Thus there is no element of $F$ that is mapped by $\phi$ into the constant function $1$.
It seems Fraleigh used a similar line of reasoning to show $\phi$ is not surjective that I did when I concluded $\phi$ is not injective. Is my answer correct too? Thanks for your help!
The function $\phi$ is injective. Indeed, if $\phi(f) = \phi(g)$, then for all $x \in {\Bbb R}$, $xf(x) = xg(x)$. It follows that, if $x \not= 0$, then $f(x) = g(x)$. But since $f$ and $g$ are continuous, $$f(0) = \lim_{x \to 0}f(x) = \lim_{x\to 0}g(x) = g(0)$$