On an injective ring homomorphism from the ring of continuous functions to the ring of differentiable functions Let $\phi : C \to D$ be an injective ring homomorphism such that $\phi(1)=1$, where $1$ denotes the constant function $1$ and $C$, $D$ are the rings of continuous, respectively differentiable functions on $\mathbb{R}$. If $f \in C ,g \in D$ are such that $\phi(f)=g$, and $t_0\in \mathbb R$, is it true that $\phi (f-g(t_0))=g-g(t_0)$ ? This is used on page 944, Problem 11617. And why if the image of $\phi $ consists of only constant functions, then $\phi$ is not injective ? Please help. Thanks in advance. 
 A: For the first part note that a standard argument shows that $\phi(q)=q$ where $q \in \mathbb{Q}$ denotes the constant function $q$. If you show that this also holds for irrational numbers you are done for the first part. I'll give a sketch of the proof.
Let $C=C(\mathbb{R})$ and $D=D(\mathbb{R})$.
Define $i: \mathbb{R} \to C(\mathbb{R})$ by $i(x) = c_x$, where $c_x$ is the constantly equal to $x$ function.
For every $x_0$ in $\mathbb{R}$ define $e_{x_0}: D(\mathbb{R}) \to \mathbb{R}$ the evaluation morphism at $x_0$.
Take $x_0$ in $\mathbb{R}$ and prove that $\psi = ev_{x_0} \circ \phi \circ i :\mathbb{R} \to \mathbb{R}$ is a ring homomorphism, which sends positive numbers to positive numbers. It may help to observe that $c_x = c_{\sqrt{x}} \cdot c_{\sqrt{x}}$ for $x > 0$.
Now $\psi:\mathbb{R} \to \mathbb{R}$ is a strictly increasing ring homomorphism which agrees with the identity on $\mathbb{Q}$. Therefore, $\psi$ is the identity. And because $x_0$ was arbitrary, then $\phi \circ i :\mathbb{R} \to D(\mathbb{R})$ maps $x \mapsto c_x$ for every $x \in \mathbb{R}$. That is, $\phi(c_x)=c_x$ for every $x\in\mathbb{R}$.
For the second part suppose that the image of $\phi$ consists only of constant functions. Take $f = \chi_{[0;1]}$ and $g = \chi_{[2;3]}$. Clearly, $f\cdot g \equiv 0$. So $0 = \phi(f\cdot g) = \phi(f)\phi(g)$. Because the image of $\phi$ consists only of constant functions, then $\phi(f) =0$ or $\phi(g)=0$. Because $f,g \neq 0$ this implies that $\phi$ is not injective, contradicting the hypothesis of $\phi$ being injective. Then, for injective $\phi$, the image of $\phi$ can't consist only of constant functions.
