Left and Right Inverses of Product

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Theorem

Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.

Let $x, y \in S$.

Let:

$(1): \quad x \circ y$ have a left inverse for $\circ$
$(2): \quad y \circ x$ have a right inverse for $\circ$.

Then both $x$ and $y$ are invertible for $\circ$.


Proof

Let $z_L$ be the left inverse of $x \circ y$ and $z_R$ be the right inverse of $y \circ x$. Then:

\(\ds z_L \circ \paren {x \circ y}\) \(=\) \(\ds e_S\)
\(\ds \leadsto \ \ \) \(\ds \paren {z_L \circ x} \circ y\) \(=\) \(\ds e_S\)


\(\ds \paren {y \circ x} \circ z_R\) \(=\) \(\ds e_S\)
\(\ds \leadsto \ \ \) \(\ds y \circ \paren {x \circ z_R}\) \(=\) \(\ds e_S\)


Thus $y$ has both a left inverse $z_L \circ x$ and a right inverse $x \circ z_R$.


From Left Inverse and Right Inverse is Inverse:

$z_L \circ x = x \circ z_R$

and $y$ has an inverse, that is, is invertible.

$\Box$


From the above, we have:

\(\ds \paren {z_L \circ x} \circ y\) \(=\) \(\ds e_S\)
\(\ds \leadsto \ \ \) \(\ds \paren {x \circ z_R} \circ y\) \(=\) \(\ds e_S\)
\(\ds \leadsto \ \ \) \(\ds x \circ \paren {z_R \circ y}\) \(=\) \(\ds e_S\)


and:

\(\ds y \circ \paren {x \circ z_R}\) \(=\) \(\ds e_S\)
\(\ds \leadsto \ \ \) \(\ds y \circ \paren {z_L \circ x}\) \(=\) \(\ds e_S\)
\(\ds \leadsto \ \ \) \(\ds \paren {y \circ z_L} \circ x\) \(=\) \(\ds e_S\)


Thus $x$ has both a left inverse $y \circ z_L$ and a right inverse $z_R \circ y$.


From Left Inverse and Right Inverse is Inverse:

$y \circ z_L = z_R \circ y$

and $x$ has an inverse, that is, is invertible.

$\blacksquare$


Sources