Question

prove that there can't be an one one map from a to b where |a|>|b|

Answer 1

f ff is not one-to-one, then there are distinct a1,a2∈Aa1,a2∈A such that f(a1)=f(a2)∈Bf(a1)=f(a2)∈B. As a result,

|f(A)|<|A|.
|f(A)|<|A|.
So there must be some b∈Bb∈B with no preimage. Hence, ff is not onto.

For the converse, just run the arguments in reverse.