This section is collection of some theorems about inverses of a funcion. Here they are, their proofs can be looked up in the book.
1. Suppose f:A -> B. f is one-to-one and onto iff :B -> A
2. Suppose f is a function from A to B and is a function from B to A. Then and
3. Suppose f:A -> B.
- If there is a function g:B -> A s.t. then f is one-to-one.
- If there is a function g:B -> A s.t. then f is onto
All of the above can be combined into this one.
4. Suppose f:A -> B. Then the following statements are equivalent.
- f is one-to-one and onto
- :B -.A
- There is a function g:B->A s.t. and
Now, this one is used in proofs alot.
5. Suppose f:A->B, If there exists a function g:B->A s.t. and . Then f is one-to-one and onto and
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Ex-1: the person sitting immediately to the right of p
Ex-2: A\X
Ex-3: Consider, g(x) = .
Then
= g(f(x))
= g()
=
= x
Thus
Also,
= f(g(x))
= f()
=
= x
Thus
It follows from theorem#5 above that f is one-to-one and onto. and = g(x) =
Ex-4: Let g be a function from to defined as g(x) =
Then,
= g(f(x))
= g()
=
=
= x
Thus
Also,
= f(g(x))
= f()
=
=
= x
Thus
Hence f is one-to-one and onto and = g(x) =
Ex-5: Scratchwork:
y =
taking log of both sides with base 10
log(y) = (2-x)log(10)
=> log(y) = 2-x
=> x = 2 - log(y)
Formal Proof:
Let defined by g(x) = 2 - log(x)
It is easy to check that and . Hence f is one-to-one and onto and = g(x) = 2-log(x)
Ex-6: Scratchwork:
y =
=> xy - 2y = 3x
=> x(y-3) = 2y
=> x =
which is not defined for y = 3, clearly B =
(a),(b) B = and consider g:B -> A defined as g(x) = .
Let x be an arbitrary element of A.
Then
= g(f(x))
= g()
=
=
=
= x
Thus
Its trivial to check that
Hence f is one-to-one and onto and = g(x) =
Ex-7:
(a) Let x be an arbitrary real number.
Then,
=
=
=
= f(x)
Since x is arbitrary, so = f
(b) Note: g means f-inverse-1 and h means f-inverse-2.
It is trivial to compute that
g(x) = x - 7
h(x) = 5x
= = = 5x-7 =
Ex-8: Prove that
(a) Let b be an arbitrary element of B. Then we can choose some s.t. f(a) = b. Then
Now, = f() = f(a) = b
Since b is arbitrary, so
(b) It is given that . Let b be an arbitrary element of B.
Then =
=> =
Thus,
Since b is arbitrary, so
Ex-9: Suppose g:A->B s.t. . Let b be an arbitrary element of B. Then . Then . Then, we can choose s.t. and . Thus f(a) = b. Since b is arbitrary, so f is onto.
Ex-10: () Let (b,a) be an arbitrary ordered pair in B X A s.t. . Since , then . Then we can choose some s.t. and . Since , and g is a function, so . Hence and then .
() Let (b,a) be an arbitrary ordered pair in B X A s.t. . Then . Since , then . Then we can choose some s.t. and . Since , and f is a function, so . Hence .
Ex-11:
(a) Since , so by theorem 5.3.3(2) f is onto. Since f is one-to-one and onto, so .
Then, g = = = = =
(b) Since , so by theorem 5.3.3(1) f is one-to-one. Since f is one-to-one and onto, so .
Then, g = = = = =
(c) Since , so by theorem 5.3.3(2) f is onto. We will prove by contradiction that f is not one-to-one. Assume f is one-to-one. Since f is one-to-one and , using part(a) g = . Since f is one-to-one and onto, so . Then , but this contradicts with the given. Hence the assumption is false and f is not one-to-one.
Since , so by theorem 5.3.3(1) g is one-to-one. We will prove by contradiction that g is not onto. Assume g is onto. Then using part(2) f = . Since g is one-to-one and onto, so . Then , but this contradicts with the given. Hence the assumption is false and g is not onto.
Ex-12: Let B' = Ran(f). We'll prove that :B' -> A.
Existence:
Let b be an arbitrary element of B'. Since B' = Ran(f), so we can choose some s.t. . Then .
Uniqueness:
Let b be an arbitrary element of B' and , be two arbitrary elements of A s.t. and . Then and . Since f is one-to-one, so .
Hence . Thus :B' -> A
Ex-13:
(a) Its trivial to see that . So f is compatible with R. By using result of ex-18(a) in section-5.1 there exists a function h:A/R -> B s.t. for all , h() = f(x)
(b) It follows straightforward from the result of ex-16, sec-5.2 that h is one-to-one.
Proof for h is onto:
Let b be an arbitrary element of B. Since f is onto, so we can choose s.t. f(a) = b. clearly , so h() = f(a) = b. Hence, h is onto.
(c) Let g(b) = {}
Let a be an arbitrary element of A. Then
= g(h()) = g(f(a)) = {} = {} = . Thus, =
Let b be an arbitrary element of B. Then
= h(g(b)) = h({})
Since f is onto, so we can choose some s.t. f(a) = b. Then
{} = {} = {} =
Then = h() = f(a) = b. Thus =
Since = and = , so (b) = g(b) = {}
NOTE: In all the exercises below, f|A denotes restriction of f to set A.
Ex-14:
(a) Let a be an arbitrary element of A'. Then we can choose some s.t. g(b) = a.
Now = = = = = g(b) = a. Since a is arbitrary, so for any ,
(b) Its easy to see that and . Hence by theorem 5.3.4, f|A' is one-to-one and onto function from A' to B. And g =
Ex-15: Let A' = Ran(g). It follows from result of ex-14(b) that g = . Also, its trivial to find that Ran(g) = B, hence A' = B. Thus g =
Ex-16:
(a) Let b be some real number s.t. f(x) = b.
Then
=>
This quadratic equation has real roots if .
Hence B = {}
(b) Scratchwork:
Using ex-14 result, basically we want to find a function g:R -> R s.t. . Then A = Ran(g) and g = .
=> f(g(x)) = x
Let g(x) = y, then
f(y) = x
=>
one root of above quadratic equation is, y = g(x) =
Formal Proof:
Let g(x) = .
= f(g(x)) = f() = = = x
Hence
It follows from ex-14 result that A = Ran(g) = {} and (x) = g(x) =
Ex-17: Scratchwork:
If we prove that and , then it follows from theorem 5.3.3 and 5.3.5 that f is one-to-one and onto and g =
Formal Proof:
Proof for :
Let R be an arbitrary element of .
Then, = g(f(R)) = g() = = R. Since R is arbitrary, so
Proof for :
Let R be an arbitrary element of .
Then, = f(g(R)) = f() =
Since R is strict partial order on A, hence irreflexive and hence R and are disjoint. Then = R. Then . Since R is arbitrary, so
Then it follows from theorem 5.3.3 and 5.3.5 that f is one-to-one and onto and g =
Ex-18:
(a) First, we prove that R is reflexive on
Let f be an arbitrary function in . Let h = . Then = f. Hence fRf. Since f is arbitrary, so R is reflexive.
Second, we prove that R is symmetric on
Let (f,g) be arbitrary ordered pair in s.t. . Then we can choose a function s.t. f = . Then . Then . Let j = . Then g = . Hence . Since (f,g) is arbitrary, so R is symmetric.
Third, we prove that R is transitive on
Let (f,g) and (g,h) be arbitrary elements of R. So we can choose functions u,v in s.t.
f = and g = .
Then f = = = . Let j = . Then f = . Hence . Hence R is transitive.
Thus R is equivalence relation on .
(b) Suppose fRg, then we can choose s.t. f = .
Then
=
=
Hence
(c) Suppose f has a fixed point and fRg. Then we can choose some s.t. f(a) = a. We can choose s.t. f = .
So f(a) =
Let h(a) = b. Then
f(a) =
=> a =
Thus
=> .
we assumed h(a) = b, that is
Since h is one-to-one, hence g(b) = b. So g has a fixed point.
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