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In An Introduction to Category Theory, exercise 1.3.6(b), Simmons writes,

Consider any pair of Pos-arrows... Show that if arrow f is inverse of arrow g, then, f(g(f)) = f && g(f(g)) = g and hence g(f) is a closure operation on A and f(g) is a co-closure operation on B.

The solution my study group got was,

Idendity of S =< f(g) && g(f) =< Identity of T
f =< f(g(f)) &&  g(f(g)) =< g
therefore f = f(g(f))

which looks like an inverse, but not quite. Is this what Simmons means by closure? Any suggestions welcome.

(I apologize that I don't know how to write tex; editors please clean up my presentation. Thanks to all in advance for the clarifications.)

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2  
    
what is the relationship between the A,B in the text and the S,T in the formulas? –  magma Jul 6 '12 at 13:58
    
I think the Wikipedia article mentioned by @QiaochuYuan answers the question. A closure of X gives you the smallest set containing X. So, g(f) gives us the smallest set containing a select element of A, and f(g) gives us the smallest set containing a select element of B. They're both closures, and more to the point, they're adjoints in the category. –  haiqus Jul 6 '12 at 22:11
    
@haiqus: I believe that $A=S$ and $B=T$, this is a mistype in the textbook. Simmons means a closure operator on a poset, not a closure operator on a powerset. So, you can not say “the smallest set containing”. $f\circ g$ is not a closure. $(f\circ g)(t)$ gives the greatest open element which is less than $t$. Either I downloaded an old copy of the textbook or you copied the exercise incorrectly. –  beroal Jul 7 '12 at 17:26
    
Yes, I see now that $A=S$ and $B=T$. He write in my copy "$g \circ f$ is a closure operation on $A$ [sic]." Is that too a typo, or am I misunderstanding something more fundamental in this exercise? –  haiqus Jul 8 '12 at 18:27

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