On page 31 of Atiyah and Macdonald, there is a commutative diagram. It essentially says that if $B$ and $C$ are $A$-algebras with ring morphisms $f:A\to B$ and $g\colon A\to C$, and $D=B\otimes_A C$ is an $A$-algebra with morphism $a\mapsto f(a)\otimes g(a)$, then $uf=vg$, where $u:B\to D$ is $u(b)=b\otimes 1$.
The map $v:C\to D$ is not defined in the text, but my guess is it's $v(c)=1\otimes c$.
I don't understand why the diagram is commutative though. That would imply $f(a)\otimes 1=1\otimes g(a)$ for all $a\in A$. Is that true, or is $v$ something else?
Added: On second thought, does this follow since $f(a)\otimes 1=a\cdot(1\otimes 1)$ and $1\otimes g(a)=a\cdot (1\otimes 1)$ where $\cdot$ is the $A$-module structure on $D$? $\require{AMScd}$ \begin{CD} A @>f>> B\\ @V g V V\# @VV u V\\ C @>>v> D \end{CD}