Suppose f : A → B and g : B → C.
1. Prove that if f is onto and g o f = i_A, then g = f^(-1).
2. Prove that if f o g = i_B but g o f ≠ i_A then f is onto but not one-to-one, and g is one-to-one but not onto.
g o f means the g composed with f, vice versa for f o g.
i_A is the identity relation of A.
i_B is the identity relation for B.
1. Prove that if f is onto and g o f = i_A, then g = f^(-1).
2. Prove that if f o g = i_B but g o f ≠ i_A then f is onto but not one-to-one, and g is one-to-one but not onto.
g o f means the g composed with f, vice versa for f o g.
i_A is the identity relation of A.
i_B is the identity relation for B.
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You must have meant to write g : B -> A---as opposed to B ->C.
1.) Suppose that f is onto and g ○ f = i_A. To show that g = f^(-1), we need to establish that f○g = i_B. To this end, let b in B be arbitrary. Since f is onto, there exists a in A such that
f(a) = b.
As g○f = i_A, we can use the fact that a = g(f(a)). Then note that
(f○g)(b) = f(g(b)) = f(g(f(a))) = f(a) = b.
Since this holds for all b in B, f○g = i_B. Combined with the hypothesis g○f = i_A, we conclude that g = f^(-1).
2.) There are four things to establish here. Throughout, we assume that f○g = i_B and that g○f ≠ i_A.
(i) f is onto: To show this, let b be any element of B. Then set a = g(b). Since f○g = i_B we have
f(a) = f(g(b)) = (f○g)(b) = b.
Thus b is in the range of f, i.e. f is onto.
(ii) g is one to one: To prove this part, let b1 and b2 be any elements of B such that
g(b1) = a = g(b2).
Evaluating f at each of these, and using the fact that f○g = i_B, we get
f(g(b1)) = f(a) = f(g(b2)) ==> b1 = f(a) = b2
that is, b1 = b2. Hence g is one to one.
(iii) f is not one to one: Here, we can use contradiction. Suppose that f is one to one. Then let a be any element of A. Set b = f(a), and note that
(g○f)(a) = g(b).
Evaluate f at the left and right side and use the fact that f(g(b)) = b
f((g○f)(a)) = f(g(b)) = b = f(a).
Since f is one to one, the above leads to
(g○f)(a) = a.
As a was arbitrary, this leads to the contradiction g○f = i_A. Hence f must not be one to one.
(iv) g is not onto: Again, we can use contradiction. Suppose that g is onto. Let a in A be arbitrary. Since g is onto, there exists b in B such that
g(b) = a.
As f○g = i_B we get
b = f(g(b)) = f(a).
Then
(g○f)(a) = g(f(a)) = g(b) = a.
As a was arbitrary, this leads to the contradictory conclusion g○f = i_A. So it must be that g is not onto.
1.) Suppose that f is onto and g ○ f = i_A. To show that g = f^(-1), we need to establish that f○g = i_B. To this end, let b in B be arbitrary. Since f is onto, there exists a in A such that
f(a) = b.
As g○f = i_A, we can use the fact that a = g(f(a)). Then note that
(f○g)(b) = f(g(b)) = f(g(f(a))) = f(a) = b.
Since this holds for all b in B, f○g = i_B. Combined with the hypothesis g○f = i_A, we conclude that g = f^(-1).
2.) There are four things to establish here. Throughout, we assume that f○g = i_B and that g○f ≠ i_A.
(i) f is onto: To show this, let b be any element of B. Then set a = g(b). Since f○g = i_B we have
f(a) = f(g(b)) = (f○g)(b) = b.
Thus b is in the range of f, i.e. f is onto.
(ii) g is one to one: To prove this part, let b1 and b2 be any elements of B such that
g(b1) = a = g(b2).
Evaluating f at each of these, and using the fact that f○g = i_B, we get
f(g(b1)) = f(a) = f(g(b2)) ==> b1 = f(a) = b2
that is, b1 = b2. Hence g is one to one.
(iii) f is not one to one: Here, we can use contradiction. Suppose that f is one to one. Then let a be any element of A. Set b = f(a), and note that
(g○f)(a) = g(b).
Evaluate f at the left and right side and use the fact that f(g(b)) = b
f((g○f)(a)) = f(g(b)) = b = f(a).
Since f is one to one, the above leads to
(g○f)(a) = a.
As a was arbitrary, this leads to the contradiction g○f = i_A. Hence f must not be one to one.
(iv) g is not onto: Again, we can use contradiction. Suppose that g is onto. Let a in A be arbitrary. Since g is onto, there exists b in B such that
g(b) = a.
As f○g = i_B we get
b = f(g(b)) = f(a).
Then
(g○f)(a) = g(f(a)) = g(b) = a.
As a was arbitrary, this leads to the contradictory conclusion g○f = i_A. So it must be that g is not onto.