Below is a proof I've written to show a ring homomorphism. This is how I was taught to write them, with the proposition, then the proof. So if it's redundant, I'm sorry. What I'm asking is whether or not my proof is accurate? Does it need more? Did I make incorrect assumptions?
Proposition: Let $F$ be the ring of functions $f: \Bbb R \rightarrow \Bbb R$, and consider the usual ring of real numbers $\Bbb R$. Define the map $\varphi_a:F\rightarrow \Bbb R$ by $\varphi_a(f)=f(a)$ for $f \in F$, then $\varphi_a: F \rightarrow \Bbb R$ is a ring homomorphism for each $a \in \Bbb R$.
Proof: Let $F$ be the ring of functions $f: \Bbb R \rightarrow \Bbb R$, and consider the usual ring of real numbers $\Bbb R$. Define the map $\varphi_a:F\rightarrow \Bbb R$ by $\varphi_a(f)=f(a)$ for $f \in F$. Let $g \in F$ such that $\varphi_a(g)=g(a)$.
Step 1: We want to show that $\varphi_a(f+g)=\varphi_a(f)+\varphi_a(g)$.
$\varphi_a(fg)=f+g(a)$
$=f(a)+g(a)$
$=\varphi_a(f)+\varphi_a(g)$
Thus, we have shown $\varphi_a(f+g)=\varphi_a(f)+\varphi_a(g)$.
Step 2: We want to show that $\varphi_a(fg)=\varphi_a(f)\varphi_a(g)$.
$\varphi_a(fg)=fg(a)$
$=f(a)g(a)$
$=\varphi_a(f)\varphi_a(g)$
Thus, $\varphi_a(fg)=\varphi_a(f)\varphi_a(g)$.
Hence, by steps 1 and 2 we have shown that $\varphi_a:F\rightarrow \Bbb R$ is a ring homomorphism for each $a \in \Bbb R$.