Answer:
False
Explanation:
Onto function:-
Let two non empty set A and B .The function is defined from set A to set B.
If the function is onto then every element of y in codomain B has at-least one pre-image in the domain A .
Or
For every element in y codomain , there is at least one element x in the domain A such that
f(x)=y
If the function is onto then
Range=Codomain
If the function is onto then every element in the co-domain musta have a pre-image not unique.
Therefore, the given statement is false.