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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.

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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.

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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

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