Answer:
See below
Step-by-step explanation:
Let S={a,b,c,d}. f : S→S is 1-1, if different elements of S have different images respect to f. f is onto if f(S)={f(x)|x∈S}=S. f is bijective if f is 1-1 and onto.
a) f is 1-1. Indeed, each element of S has a different image respect to f, thus no different elements of S have the same image (f(x)≠f(y) if x≠y).
f is onto, because f(S)={f(a),f(b),f(c),f(d)}={b,a,c,d}={a,b,c,d}=S. Then the direct image of f is equal to the codomain S, hence f is onto.
f is 1-1 and onto, hence f is bijective.
b) f is not 1-1. a≠b but f(a)=b=f(b).
f is not onto. a∈S, but there is not element x such that f(x)=a. In other words, the direct image f(S)={f(a),f(b),f(c),f(d)}={b,b,c,d}={b,c,d} is not equal to S.
f is not bijective, since f is not 1-1.
c) f is not 1-1. a≠d but f(a)=d=f(d).
f is not onto. a∈S, but there is not element x such that f(x)=a. In other words, the direct image f(S)={f(a),f(b),f(c),f(d)}={d,b,c,d}={b,c,d} is not equal to S.
f is not bijective, since f is not onto.
