Is $\phi: C^{\infty}(\mathbb{R}) \to (\mathbb{R},+), \phi(f) = f'(0)$ an isomorphism?

Question

Let $F$ be the set of all functions $f: \mathbb{R} \to \mathbb{R}$ that have derivatives of all orders.

Is $\phi$ an isomorphism from the first structure to the second where,

$\phi:(F,+)\to(\mathbb{R},+)$ and $\phi(f) = f'(0)$ ?

1) one-to-one

let $f,g \in F$ where $f=e^x$ and $g=\sin x$

now, $\phi(e^x) = 1$ and $\phi(\sin x) = 1$ but $e^x \neq \sin x$ thus it is not a one-to-one mapping and so $\phi$ is not an isomorphism.

Answer 1

Your argumentation is right. You can also say, that for each $\alpha \in \mathbb R\setminus\{0\}$ and $f \in F$ both functions $f$ and $f+\alpha$ are mapped to the same value via $\phi$.