Bilinear Morphism of Monoids

A function $f:X\times Y\to Z$ from the Product Monoid $X\times Y$ to a Monoid $Z$ is bilinear if it is a Monoid Homomorphism "separately" in each variable. Explicitly, this means we have:

• $f(0,y)=0$
• $f(x,0)=0$
• $f(x_{1}+x_{2},y)=f(x_{1},y)+f(x_{2},y)$
• $f(x,y_{1}+y_{2})=f(x,y_{1})+f(x,y_{2})$

We have written this in additive notation as $X,Y$ and $Z$ are typically Commutative, but the definition works for arbitrary monoids

Properties

• If $f:X\times Y\to Z$ is a bilinear morphism, then so is $f\circ\sigma:Y\times X\to Z$. More generally, if $f:\prod_{i:I}X_{i}\to Z$ is a Multilinear Morphism of Monoids, then precomposing by a Permutation $\sigma:I\simeq I$ yields another Multilinear Morphism.

• A bilinear morphism of monoids $f:X\times Y\to Z$ preserves all Invertible Elements separately in each variable.

  Concretely, suppose that we have some $x^{-1}:X$ with $x\cdot x^{-1}=1=x^{-1}\cdot x$. Our claim is that for all $y:Y$, $f(x^{-1},y)$ is an inverse to $f(x,y)$.

  This follows from a pair of short calculations

  	$\displaystyle f(x,y)\cdot f(x^{-1},y)$	$\displaystyle=f(x\cdot x^{-1},y)$
  		$\displaystyle=f(1,y)$
  		$\displaystyle=1$

  	$\displaystyle f(x^{-1},y)\cdot f(x^{,}y)$	$\displaystyle=f(x^{-1}\cdot x,y)$
  		$\displaystyle=f(1,y)$
  		$\displaystyle=1$

  We can use the fact that bilinear morphisms are stable under permutations to deduce the other case.

• If $X$ and $Z$ are Groups, then we can omit