Being $A, B$ and $C$ rings. To prove the above statement, I need to show that $ ψ \circ φ: A → C $ satisfies the following conditions:
i) $ψ \circ φ (a + b) = ψ \circ φ (a) + ψ \circ φ (b)$ $\forall$ $a,b \in A$
ii) $ψ \circ φ (ab) = ψ \circ φ (a) ψ \circ φ (b)$ $\forall$ $a,b \in A$
This is my thinking about this demonstration, but I do not know how to do it.
Let $a, b \in A$. Assume $ψ$ and $φ$ homomorphisms.
$ψ \circ φ(a + b)=ψ(φ(a+b))=ψ(φ(a)+φ(b))=ψ(φ(a))+ψ(φ(b))=ψ \circ φ(a)+ψ \circ φ(b)$
and
$ψ \circ φ(ab)=ψ(φ(ab))=ψ(φ(a)φ(b))=ψ(φ(a))ψ(φ(b))=ψ \circ φ(a)ψ \circ φ(b)$
Could you tell me if this approach is correct?