Suppose $F$ is the set of all functions $f$ mapping $\mathbb{R}$ into $\mathbb{R}$ that have derivatives of all order. Then is $\phi$ an isomorphism of the first binary operation with the second?
- $
$ with $$ where $\phi(f) = \int_{0}^{x}f'(t) dt$
No, because $\phi$ does not map $F$ onto $F$. For all $f\in F$, we see that $\phi(f)(0)=0$ so, for example, no function is mapped by $\phi$ into $x+1$.
I am not sure what the solution means.
Is it saying that if $x = 0$, then $\int_{0}^{0}f(t)dt = 0 \implies \phi(x+1)(0) = \int_{0}^{0}t+1 dt = 0$? Because that is true. So why isn't it onto?