Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A, B, C$ be nonempty sets with total order. and let $f\colon A \rightarrow B$ and $g\colon B \rightarrow C$ be maps. Prove these statements:

a) If $f, g$ are antitone, then $g \circ f$ is isotone
b) If $f, g$ are strictly isotone, then $g \circ f$ is injective
c) If $f, g$ are injective, then $g \circ f$ strictly is isotone

I am stuck now not knowing where and with what to start. I started like this:

$$f\text{ is antitone} \quad:\Longleftrightarrow\quad x<y \Rightarrow f(x) \geq f(y)$$
$$g\text{ is antitone} \quad:\Longleftrightarrow\quad x<y \Rightarrow g(x) \geq g(y)$$

... I don't know how to show that $g \circ f$ is isotone. :(

Can someone help me please with all these?

Thanks a lot.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

All you have to do is plug in the immediate definitions:

a) Let $x,y\in A$ with $x\le y$ be given. Then $f(y)\le f(x)$ because $f$ is antitone and thus $g(f(x))\le g(f(y))$ because $g$ is antotone. Thus $x\le y$ implies $(g\circ f)(x)\le (g\circ f)(y)$, which is the very definition of $g\circ f$ is isotone".

b) You may already know that the condition implies that $g\circ f $ is strictly isotone. And that every strictly isotone map is injective.

c) I assume you are rather expected to exhibit a simple counterexample instead of proviing this obviously wrong statement? Let $A=B=C=\{0,1,2\}$, $f(x)=x+1\mod 3$, $g(x)=x$. Then both $f$ and $g$ are injective, but their composition (which is just $f$) is neither iso- nor antitone.

share|improve this answer
    
WOW, thanks Hagen, great! –  doniyor Nov 17 '12 at 10:34
    
Hagen, i am coming back to your answer. we have here $g: B \rightarrow C$. why we say $g(f(x))\le g(f(y))$? this has nothing to do with our original $g$ function. :( –  doniyor Feb 3 '13 at 9:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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