Is the implication $(P \circ Q = \mathrm{id}) \wedge (Q \circ P = \mathrm{id}) \rightarrow P = Q^\dagger$ true in every allegory? In the allegory of sets and relations, we have the following (remarkable?) fact: if two relations $P$ and $Q$ are inverses, meaning $P \circ Q = \mathrm{id}$ and $Q \circ P = \mathrm{id}$, then it's also the case that $P = Q^\dagger$. This kind of thing doesn't work in most dagger categories. For example, it's not the case that if $A$ and $B$ are inverse matrices, then $A = B^\top$. > **Question.** Is $$(P \circ Q = \mathrm{id}) \wedge (Q \circ P = \mathrm{id}) \rightarrow P = Q^\dagger$$ true in every allegory? If not, in which allegories is it true?

$\def\d{^\dagger}$ **Yes** , $P^{-1}=P\d$ holds in every allegory, using only $P\le PP\d P$.

For one part, $PQ=1$ implies $$1=PQ\le PP\d PQ=PP\d\\\ 1=PQ\le PQQ\d Q=Q\d Q$$ We obtain dual inequalities for $QP=1$.

The other direction of inequalities is a bit longer and uses both $PQ=1$ and $QP=1$ at once. $$P\d P = P\d PQP\ \le\ P\d PQQ\d QP=P\d Q\d =1\d=1\,.$$