im having trouble on a problem
A={1,2,3,4,5,6} f:A->A
i need to find an example where f is bijective, but not 1A
I don't understand this statement. f: A -> A doesn't require that A be the image but only the codomain, so for example the function $\displaystyle f(x) = 1 \;\forall x \in A$ would conform to f: A -> A.
But I also don't see the difficulty in finding a bijective function. Just write down
$\displaystyle f(1) = \boxed{ }$
$\displaystyle f(2) = \boxed{ }$
...
$\displaystyle f(6) = \boxed{ }$
and fill in with values from A such that you use every single element, then it will be bijective. (Sorry the rectangles are a bit uncentered vertically, couldn't find the proper LaTeX.)
The word `into' is sometimes used to mean `injection' (much like `onto'). When there exists an injection, f, from A to B then it means you can find a copy of A in B, and so f maps A into B.
I would also presume, as did undefined, that 1A means the identity function ($\displaystyle 1_A$ or $\displaystyle I_A$) as this would be too easy an answer and noone would exclude any other bijection.
Now, bijections between finite sets are just permutations. So, permute your elements. For example,
$\displaystyle 1 \mapsto 2$
$\displaystyle 2 \mapsto 1$
$\displaystyle 3 \mapsto 3$
$\displaystyle 4 \mapsto 4$
$\displaystyle 5 \mapsto 5$
$\displaystyle 6 \mapsto 6$
will be a function which switches 1 and 2. It is thus a permutation, as it permutes these two elements and keeps everything else fixed.