Definitions:
$1)$ $R$-Algebra: Let $R$ be a commutative ring with $1\neq 0$. An $R$-Algebra is a ring $A$ with $1\neq 0$ together with a ring homomorphism $f:R\rightarrow A$ such that $f(1_{R})=1_{A}$.
$2)$ $R$-Algebra Homomorphism: Let $A$ and $B$ be two $R$-Algebras. An $R$-Algebra Homomorphism is a ring homomorphism $\varphi:A\rightarrow B$ mapping $1_{A}$ to $1_{B}$ such that $\varphi (r.a)=r.\varphi (a)$ $\forall r\in R$ and $a\in A$.
My Question:
Let the functions associated with $A$ and $B$ $R$-Algebras be $f_{1}$ and $f_{2}$ respectively. Then, does the diagram below commutes i.e. $\varphi \omicron f_{1}= f_{2}?$
$\require{AMScd}$
\begin{CD} R&\xrightarrow{f_{1}} &A\\ & f_{2}\searrow &\downarrow{\varphi}\\& &B\end{CD}
My intuition says it is not necessary. But I have no proof or counterexample to prove it. Moreover, I want to know whether by imposing some conditions on $f_{1}$, $\varphi$ and $f_{2}$, can we make the above diagram commutative?
Best Answer
For $r \in R, a \in A$ we have by definition $r.a = f_1(r) a$ and for $b \in B$ we have $r.b = f_2(r) b$. Now $$\varphi(f_1(r)) = \varphi(r . 1_A) = r. \varphi(1_A) = f_2(r) 1_B = f_2(r).$$