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The question speaks for itself, almost. But to clarify the context, we are working in a ring $A$, and have that $ab$ and $ba$ are invertible. Then there are $c, d ∈ A$ such that $b(ad) = 1 = (ca)b$. It should follow that both $b$ and $a$ are now invertible, but since I cannot write down $b^{-1}$ or $a^{-1}$, I can't tie a bow on the argument.

Jos van Nieuwman
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    Hint. Suppose an element $x$ has a left inverse $l$ and a right inverse $r$. So the only elements in your ring you know something about (besides $0$ and $1$) are those three. Now three elements suggests to test associatlvity... – Andreas Caranti Mar 22 '22 at 15:49
  • Omg I'm an idiot :p – Jos van Nieuwman Mar 22 '22 at 15:50
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    Please include the question in the body of your post, and not only in the subject line. The subject line is like the address on an envelope: you don't start the letter on the envelope. – Arturo Magidin Mar 22 '22 at 17:02
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    We can view this method of proof hinted by @Andreas as a special case of "overlapping" (unification) of terms to derive consequences (a general method used in many rewriting algorithms used to derive equational consequences), as I explain here. That viewpoint is helpful in making analogous deductions in many common proofs. – Bill Dubuque Mar 22 '22 at 17:40

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A curious fact: this holds in much more generality.

The following might not be of that much help if you do not know any category theory though. Anyway, fix a category $\mathcal C$ and a morphism $f\colon x\to y$ in $\mathcal C$. If there are $g\colon y\to x$ and $h\colon y\to x$ such that $g\circ f=1_x$ and $f\circ h=1_y$, then $g=h$. In particular, $f$ is an isomorphism.

This is a straighforward application of assocaitivity of composition and the identity laws:

$$ g=g\circ 1_y=g\circ(f\circ h)=(g\circ f)\circ h=1_x\circ h=h $$

Why do I bring this up? Well, this formal proof captures the essence of the argument. For example, the ring structure is mostly irrelevant as we only work in the multiplicative monoid. And a monoid is nothing else than a category with one object, although this is not important here. But you can see that the generalization is (very) natural.

In your case, the argument reads as

$$ ca=(ca)1=(ca)(b(ad))=((ca)b)(ad)=1(ad)=ad $$

proving that $b$ has a unique inverse $ca=ad$. A similar argument applies to $a$.

mrtaurho
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