Simple Isomorphism confusion. 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?


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*$<F,+>$ with $<F,+>$ 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?

For every function in $\mathrm{Im}(\phi)$, the image of $0$ is $0$.
There are many functions which are infinitely differentiable which are not $0$ when evaluated at $0$.
Therefore $\mathrm{Im}(\phi)\neq F$.
