Let $\text{Pos}$ be the category of partially ordered sets and monotonic functions. A morphism $f$ is called an isomorphism if there is a morphism $g$ such that $f\circ g$ and $g\circ f$ are identity morphisms. On the other hand a bijective morphism satisfies this condition,while it has been asserted that they are not identical. Can you explain what is happening? thanks in advance.


Take two elements, $0$ and $1$, and define the poset $X$ with $0\leq 1$ (and the reflexivity conditions), and $Y$ the poset only with $0\leq 0$ and $1\leq 1$ (i.e., we can't compare $0$ and $1$ in $Y$. Then the identity $Y\to X$ is a bijective morphism, but not an isomorphism (what would be it's inverse, and why is it not a morphism?).

The problem with posets is that we can't always compare elements.

If we were in the category of totally ordered sets, then any bijective morphism is in fact an isomorphism. What We did above was just take a simple poset ($X$) and weaken its structure a little bit (and obtain $Y$).

  • $\begingroup$ Thanks. my problem lied in the wrong hypothesis that all monotonic functions were morphisms. $\endgroup$ – user318848 Feb 29 '16 at 22:09
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    $\begingroup$ @user318848 Of course every monotonic functions is a morphisms of $\mathrm{Pos}$ : this is the definition of a morphism in the category of posets. Your error lied in the belief that the set-theoretic inverse of a monotonic function is necessarily a monotonic function. $\endgroup$ – Pece Mar 1 '16 at 10:17
  • $\begingroup$ @Pece That was exactly what I falsely believed. $\endgroup$ – user318848 Mar 1 '16 at 20:15

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