Question

Determine whether or not the given map is an isomorphism on the structures . If it isn’t explain why. Let F be the set of all functions f mapping R —> R that have derivatives of all orders . <F,+> with <R,+) with phi (f) = f’(0) for f is an element of F

Answer 1

Consider the structures $\langle F,+\rangle$ and $\langle \mathbb R,+\rangle$, and the map $\psi: F\to \mathbb R$, $\psi (f) = f'(0)$. This map is not injective. Indeed, for the different functions $f_1(x)=2e^x$ and $f_2(x)=e^{2x}$ that have derivatives of all orders, in particular, $f_1'(x)=2e^x$ and $f_2'(x)=2e^{2x}$, we have that $\psi (f_1) = f_1'(0)=2e^0=2$ and $\psi (f_2) = f_2'(0)=2e^{2\cdot 0}=2.$ Therefore, $\psi (f_1) =\psi (f_2)$, and $\psi$ is not a injection. Consequently, $\psi$ is not an isomorphism.