Extension of Scalars

Let $M$ be a $A,B$ Semiring Bimodule, $f:A\to R$ , $g:B\to S$ be a pair of Semiring Homomorphisms, as in the following diagram:

The extension of scalars of $M$ along $f,g$ is the $(R,S)$ Semiring Bimodule $f_{!}Mg_{!}$ formed by taking the Tensor Product

$\par mathrm{id}{ifx..lse\par i}^{*}Rf^{*}\otimes_{A}M\otimes_{B}g^{*}Smathrm{% id}{ifx..lse\par i}^{*}$

where

$mathrm{id}{ifx..lse\par i}^{*}Rf^{*}$ is the Restriction of Scalars of $R$ regarded as an $(R,R)$ bimodule along the pair $(mathrm{id}{ifx..lse\par i},f)$ .
$g^{*}Smathrm{id}{ifx..lse\par i}^{*}$ is the Restriction of Scalars of $S$ regarded as an $(S,S)$ bimodule along the pair $(g,mathrm{id}{ifx..lse\par i})$ .

Moreover, there is an $(f,g)$ Semilinear Map $\varphi:M\to f_{!}Mg_{!}$ defined to be $\varphi(x)=(1,x,1)$ ; this is clearly a Monoid Homomorphism, as $(1,x+y,1)=(1,x,1)+(1,y,1)$ in the tensor product of modules. Finally, we have

$\displaystyle(1,a\cdot x\cdot b,1)$	$\displaystyle=(1\cdot a,x,b\cdot 1)$	(defn. of tensor product)
	$\displaystyle=(1f(a),x,g(b)1)$	(by defn.)
	$\displaystyle=(f(a)1,x,1g(b))$	(left/right identities)
	$\displaystyle=f(a)\cdot(1,x,1)\cdot g(b)$	(by defn.)

which shows that $\varphi$ is indeed semilinear. This fills in the missing portion of our square:

Moreover, the extension of scalars is the universal such filler; for any other squares

there exists a unique factorization of $\psi$ through $\varphi$ , as below

In much more consise terms, the extension of scalars is the Cocartesian Lift of

M

along

(f,g)

in the Double Category of Bimodules.