We have set A with m elements and set B with n elements. If m<n, Find the number of functions from A onto B.
Okay so a Function is A-->B is a relation f:A-->B then 1.) Domf=A, 2.) Every x in DomA has unique image in B. (if (x,y) in f and (x,z) in f then y=z)
And onto is if Rngf=B (every image in B has a preimage in A)
Now i did an example with A={a,b} and B={c,d,e} so m<n.
i used the string approach
Set B: c d e
Choices from set A: 2 1 0
Apply the product rules is 1*2*0=0 so we have 0 onto fns from A to B
Now to prove for all numbers.
set A={a(1), a(2)....a(m)}
set B={a(1),a(2)...a(m)...a(n)}
So m<n
so with strings
set B: a(1) a(2)... a(m)...(an)
# choices from set A: a(n) a(n-1)..a(m)....0
so applying the product rule a(n)*a(n-1)*...*a(m)*...0=0
so if m<n then there can be 0 onto functions.
Is this wrong? My teacher says its more complicated but i think hes being complicated. Please any feedback would be gracious.
My teacher told me this:
This problems is more involved than that. The easiest way to solve it is by counting the number of non-onto functions (missing 1, 2, 3, ...n target elements) and then conclude that # onto functions=#functions-# non-onto functions. Get a feeling for this by working with |A|=5, |B|=4: 1. Count the number of functions from A to B 2. Count the number of non-onto functions from A to B Number of functions missing one target + # functions missing two targets +# functions missing 3 targets + # functions missing 4 targets 3. Now count # onto functions
Yep. This is even true for infinite sets, since |X| < |Y| ::= "There is an injection from X to Y but no bijection." The only way that can happen is if there is no possible surjection from Y to X.
(Notably, the converse does not hold in absence of choice: a surjection from X to Y does not necessarily imply a injection exists from Y to X, at least not in ZF~C.)