Functions proof

[jaunty fjord]

For every non empty set A, if f is a one-to-one function and onto function from A to A, then f o f =IA (identity A).
I am stuck becuase i said its false but im not sure if I can choose the function. I said chose A={1,2,3} but can i then say make f={(1,2),(2,1),(3,3)}?

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[cosmic granite]

f^2 there is identity

but yes, take $f:(1,2,3)\mapsto(2,3,1)$

[open ember]

Omegabet_

[cosmic granite]

(if you know permutations, f is the cycle (123))

[jaunty fjord]

ok so i can choose f therfore making the stament false

[cosmic granite]

yes, take A={1,2,3} and have f act like that

[jaunty fjord]

ok thanks

[cosmic granite]

then from computation it's clear $(f\circ f)(1)=f(2)=3\neq 1$

[open ember]

Omegabet_

[jaunty fjord]

prefect thanks so much

+close