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I am having trouble with the following lemma from J. Rotman's Galois Theory book:

Lemma $83$. If $L$ and $L'$ are lattices and $\gamma: L \rightarrow L'$ is an order reversing bijection $[a \leq b$ implies $\gamma(b) \leq \gamma(a)]$, then $\gamma(a \wedge b) = \gamma(a) \vee \gamma(b)$ and $\gamma(a \wedge b) = \gamma(a) \vee \gamma(b)$.

The proof is fairly obvious if you know that when $\gamma$ is order reversing, then $\gamma^{-1}$ is order reversing also. Rotman says that this is "easily seen", but I just don't see it. I think it would be easy if his definition was that $\gamma$ is order reversing when $a \leq b$ if and only if $\gamma(b) \leq \gamma(a)$ instead. In a set theory book I have, isomorphisms between two posets $A$ and $B$ are defined as $f: A \rightarrow B$ such that $a \leq b$ if and only if $f(a) \leq f(b)$.

Am I missing something really easy or should the definition be different? Is there a counterexample? That is, is there a bijective map $f: A \rightarrow B$ between two lattices/posets $A$ and $B$ such that $a \leq b$ implies $f(b) \leq f(a)$, but $f^{-1}$ does not have this property?

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Try to break it down: first prove this is true for $L^{op}$ which is $L$ with the reversed order (prove that join and meet are reversed too), then prove the lemma by showing that an order-reversing bijection is an isomorphism of $L$ with $(L')^{op}$. –  Asaf Karagila Jul 1 '12 at 23:45
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2 Answers 2

up vote 4 down vote accepted

This is not true. If you have the lattices $L=\{\bot,a,b,\top\}$ (with $\bot<a,b<\top$) and $L'=\{0,1,2,3\}$ with the usual order, the order-preserving map $$f(\top)=3\\ f(a)=1\qquad f(b)=2\\ f(\bot)=0$$ is of course a bijection, but $1<2$ yet $f^{-1}(1)=a$ and $f^{-1}(2)=b$ are incomparable.

In other words, $f$ needs to satisfy [$a\le b$ iff $f(a)\le f(b)$] to be an isomorphism for the order structure (indeed, $L$ and $L'$ are not isomorphic).

And this is also a necessary condition for the conclusion of your lemma, since when the order-preserving analogue of the lemma holds, $f(a)\le f(b)$ implies $f(a)=f(a)\wedge f(b)=f(a\wedge b)$ so that because $f$ is injective $a\le b$.


Edit: As said in the comments, when $L$ and $L'$ are isomorphic and finite, it is however true that any bijective $f$ satisfying [$a\le b$ implies $f(a)\le f(b)$] is an isomorphism between $L$ and $L'$. This is because the image of the relation $<_L$ is a subset of $<_{L'}$; but they both are finite sets of equal cardinal so they must be equal.

In fact we have this beautiful generalization of the well-known statement about finite sets:

An injection between two finite sets of equal cardinal is a bijection.

to finite posets:

An order-preserving injection between two finite posets of equal order types is an isomorphism.

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Good job! I didn't read the definition of lattices before writing my answer so I didn't know you needed a join and a meet for everyone, but you still managed to find a counter example under that constraint. The big idea is that there exists "posets-homomophisms" (i.e. order-preserving maps) which are bijective onto some poset which has less structure. +1 –  Patrick Da Silva Jul 2 '12 at 5:38
    
Awesome, thanks!! Taking $f(\top) = 0$, $f(a) = 1$, $f(b) = 2$, $f(\bot) = 3$ gives us the counterexample for order-reversing maps. –  spin Jul 2 '12 at 10:28
    
@PatrickDaSilva: Actually $L'$ has more structure than $L$ (the image of the relation $<_L$ is a proper subset of $<_{L'}$), and in fact the existence of an order-preserving bijection tells us that $|<_L|\le|<_{L'}|$. If the bijection is not an isomorphism and the sets are finite, the inequality is strict, so that if $L$ and $L'$ are finite and isomorphic, all order-preserving bijections are actually isomorphisms. –  Generic Human Jul 2 '12 at 15:07
    
@Generic Human : Perhaps I read things upside down (I still have problems with left and right, so...) but I think you see my point. –  Patrick Da Silva Jul 2 '12 at 15:20
    
I do :-) It just made me notice that there is a special case where the claim in the question happens to be true; really my remark belongs more in the answer itself, so I added it. –  Generic Human Jul 2 '12 at 18:31
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You need to assume the if and only if statement. As a counter example, just consider $L$ as the poset with an order that "can't compare distinct elements at all" (i.e, there are no elements $a \neq b$ for which $a \le b$ or $b \le a$) and clearly any bijection to another poset will satisfy the conditions on your $\gamma : L \to L'$ above, but you can't possibly expect it to be order reversing, or that $\gamma^{-1}$ has the same property. But then again, I don't think that in that case you can expect that $\gamma( a \vee b) = \gamma(a) \wedge \gamma(b)$, and so on, since the condition is satisfied trivially...

Hope that helps,

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But in lattices, this doesn't work so well? If we have the $L$ you describe and $a \neq b$, then $a \wedge b \leq a \leq a \vee b$ so $a \wedge b = a = a \vee b$ and thus $a \leq b \leq a$ and $a = b$, a contradiction. Thus any such lattice $L$ would necessarily have only one element.. –  spin Jul 1 '12 at 23:55
    
@spin : I must admit I didn't recall any property of lattices. I should've read the definition, perhaps. Thanks for giving me the tip. So my counter example is bad, but the idea is good though. Look at Generic Human's proof =) –  Patrick Da Silva Jul 2 '12 at 5:34
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