Consider the Cartesian product AB where AB are finite nonemp

Consider the Cartesian product A×B, where A,B are finite nonempty sets,
each with cardinality greater than 1. There are two functions with domain
A×B, called projections, with mapping rules p1(a,b) = a and p2(a,b) = b.
What is the target space of p1? Of p2? Are either of p1, p2 one-to-one?
Onto?

Solution

Target space is basically range or the output of a function

So it will be a for p1 and b for p2

A function f:ABf:AB is called one-to-oneone-to-one or injectiveinjective if the function ff is injective if f(x)=f(y)x=yf(x)=f(y)x=y. In words, the function ff is injective if each distinct element of AA (x,yA,xy)(x,yA,xy) gets mapped to a different element of BB. If two distinct elements of AA get sent to the same element of BB under ff, then ff is not injective (one-to-one).

A function f:ABf:AB is called ontoonto or surjectivesurjective if for every element bBbB, we can find an element aAaA such that b=f(a)b=f(a). In words, for every element bBbB in the codomain (or target space, as you call it) there is an element aAaA in the domain that maps to bb under ff. If we can\'t find such an aa for every element bb in the codomain BB, then ff is not surjective (onto).

For your first function, p1:A×BAp1:A×BA, (a,b)a(a,b)a, do the distinct arbitrary ordered pairs (a1,b1),(a2,b2)A×B(a1,b1),(a2,b2)A×B map to the same element in AA? (Note that \"distinct\" in this case means that (a1a2,b1b2)(a1a2,b1b2). For every aAaA, can we find an ordered pair such that f(a,b)=af(a,b)=a? You can apply the definitions directly here. For the function p2p2, it\'s exactly the same.

According to this both are onto functions