Algebras - x Has Quasi-Inv. y IFF (0, 1) - (x, 0) Has Inverse (0, 1) - (y, 0)

# Algebras - x Has Quasi-Inverse y IFF (0, 1) - (x, 0) Has Inverse (0, 1) - (y, 0)

 Proposition 1: Let $\mathfrak{A}$ be an algebra and let $x, y \in \mathfrak{A}$. Then $x$ has quasi-inverse $y$ if and only if in the unitization $\mathfrak{A} + \mathbf{F}$ we have that $(0, 1) - (x, 0)$ has inverse $(0, 1) - (y, 0)$.

Note that in the above proposition we do NOT require $\mathfrak{A}$ to have a unit. This is one reason why the unitization $\mathfrak{A} + \mathbf{F}$ is significant because the unitization is always an algebra with unit $(0, 1)$.

• Proof: Let $f : \mathfrak{A} \to \mathfrak{A} + \mathbf{F}$ be defined for all $x \in \mathfrak{A}$ by $f(x) = (x, 0)$. Then $f$ is easily checked to be a monomorphism (and injective homomorphism).
• Let $x, y \in \mathfrak{A}$.
• $\Rightarrow$ Suppose that $x$ has quasi-inverse $y$. Then $x \circ y = y \circ x = 0$. Now observe that:
(1)
\begin{align} \quad \quad (0, 0) \overset{(*)} = f(0) = f(x \circ y) = f(x + y - xy) = (x + y - yx, 0) = (x, 0) + (y, 0) - (x, 0)(y, 0) = (x, 0) \circ (y, 0) \end{align}
• And similarly:
(2)
\begin{align} \quad \quad (0, 0) \overset{(*)}= f(0) = f(y \circ x) = f(y + x - yx) = (y + x - yx, 0) = (y, 0) + (x, 0) - (y, 0)(x, 0) = (y, 0) \circ (x, 0) \end{align}
• (Where the equalities at $(*)$ come from the fact that $f$ is a monomorphism).
• The above two equations show that $(x, 0)$ is quasi-invertible with quasi-inverse $(y, 0)$. Since $X + \mathfrak{A} + \mathbf{F}$) that $(0, 1) - (x, 0)$ is invertible with inverse $(0, 1) - (y, 0)$.
• Now observe that:
(3)
\begin{align} \quad \quad f(0) = (0, 0) \overset{(**)} = (x, 0) \circ (y, 0) = (x, 0) + (y, 0) - (x, 0)(y, 0) =(x + y, 0) - (xy, 0) = (x + y - xy, 0) = f(x + y - xy) \end{align}
• (Where the equality at $(**)$ comes from the fact that $(x, 0)$ is quasi-invertible with quasi-inverse $(y, 0)$).
• So $f(0) = f(x + y - xy)$. Since $f$ is injective (as it is a monomorphism), $0 = x + y - xy = x \circ y$. A similar argument shows that $0 = y \circ x$. So $x$ is quasi-invertible with quasi-inverse $y$. $\blacksquare$