I'm having trouble understanding isomorphisms.

E.g. in the category Posets which Awodey defines to be the category with posets as objects and monotone functions as arrows, he explains that bijective homomorphisms are not necessarily isomorphisms. His example are the Posets $A=(U,≤_A)$ and $B=(U,≤_B)$ with $U=\{0,1\}$ and $≤_A=\{(0,0),(1,1)\}$ and $≤_B=\{(0,0),(0,1),(1,1)\}$.

Now there is obviously an Arrow $f:A→B$ corresponding to the identity on U and the identity on U is also an arrow $g:B→A$. What I don't understand is: why are they not inverse?

Of course it seems to be like this because B is totally ordered while A is not, but I don't see how I can prove that. If I define arrows to be monotone functions they operate on Elements in U and for those $f\circ g$ is the identity.

  • 2
    $\begingroup$ Note that isomorphisms of posets are bijective maps $f : P \to Q$ which satisfy $p \leq p'$ if and only if $f(p) \leq f(p')$. (Similarly, isomorphisms of topological spaces are bijective maps $f:X \to Y$ such that $U \subseteq Y$ is open if and only if $f^{-1}(U)$ is open.) $\endgroup$ – Martin Brandenburg Nov 23 '14 at 21:36
  • $\begingroup$ That makes sense. Thanks for your explanation. $\endgroup$ – cocreature Nov 23 '14 at 21:43

You have $0\le_B 1$, but $g(0)\not\le_A g(1)$. So $g$ is not an arrow in $\mathbf{Poset}$.

More generally, $(U,\le_A)$ is totally disordered, so any map $U\to V$ ($V$ with any order relation) is monotone, because there's no pair that can falsify it.

It's quite easy to see that if $A=(U,\le_A)$ and $B=(V,\le_B)$ are isomorphic in $\mathbf{Poset}$ and $A$ is totally disordered, then also $B$ is. With a similar technique, if $A$ is totally ordered, then any poset isomorphic to $A$ is again totally ordered.

As Martin Brandenburg observes, $f\colon A\to B$ is an isomorphism if and only if

  • $f$ is bijective and
  • $x\le_A y$ if and only if $f(x)\le_B f(y)$.
  • $\begingroup$ How did I not see that? Thx! $\endgroup$ – cocreature Nov 23 '14 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.