Prove: A mapping f:S->T is bijective if and only if it has an inverse?

In summary: Can a mapping from f:S->T associate an element of s into several elements of T? Also, how do you prove: A mapping f:S->T is bijective if and only if it has an inverse?Yes, a mapping from f:S->T is bijective if and only if it has an inverse. To prove this, first note that for any y in T, let x= f-1(y). Then, by definition of inverse, f(x)= y so f is surjective. Since f is injective, there is only one such. Define g: T->S by g(y)= x. Now show that that is a function and is
  • #1
mruncleramos
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Can a mapping from f:S->T associate an element of s into several elements of T? Also, how do you prove: A mapping f:S->T is bijective if and only if it has an inverse?
 
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  • #2
No. "mapping" is another word for "function". Part of the definition of function is that f(x) be a single value.

f:S-> T is bijective iplies f has an inverse:

Let y be any member of T. Since f is bijective it is both one-to-one (injective) and onto (surjective). Since f is surjective, there exist at least one member of S, x, such that f(x)= y. Since f is injective, there is only one such. Define g: T->S by g(y)= x.
Now show that that is a function and is the inverse of f.

f has an inverse implies f is bijective.

i) for any y in T, let x= f-1(y). Then, by definition of inverse, f(x)= y so f is surjective.

ii) suppose f(x1)= f(x2). Applying f-1 to both sides of the equation f-1(f(x1)= f-1(x2)
x1= x2
so f is injective.

Since f is both injective and surjective, it is bijective.
 
  • #3
mruncleramos said:
Can a mapping from f:S->T associate an element of s into several elements of T? Also, how do you prove: A mapping f:S->T is bijective if and only if it has an inverse?

To take things into an account, I remember in discrete maths - a bit like computational mathematics or rather, it is computational mathematics - we define set |R| = |{1,1,1,1}| as 4. As there are four items in the set. Now, in the other extreme, let S = {1}, then |S| = |{1}| = 1. So in effect the sets R and S have different properties.

One other thing is that, once again, in computer science, R = {1,1,1,1} and S = {1} are two totally different sets. Let's map (+2) to both sides and find the total sum of all the elements in the sets, i.e., {sum (map (+2) R)} and {sum (map (+2) S)}.

sum (map (+2) R) = sum {3,3,3,3} = 12
sum (map (+2) S) = sum {1} = 1 /= 12 = sum (map (+2) R)
and as function applied two both sets are equal, R and S are not equal.

However, answering your question, even in computer science the values of each element after mapping only takes 1 value from the image. i.e., notice that (map (+2) R) = {3,3,3,3}. So 1 -> 3 by mapping. And each element valued 1, goes to exactly one value in the image of f, that is 3.

But as I said, this is in computer science and most of the computer applications - although duplicates are considered nasty in some cases.
 
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  • #4
The set {1,1,1,1} contains only one element. you aren't allowed to repeat elements in a set.
 
  • #5
I meant cardinalities... yes my wordings were pretty poor.
 

1. How do you define a mapping?

A mapping, also known as a function, is a relation between two sets where each element in the first set (domain) is paired with exactly one element in the second set (codomain). The mapping is denoted as f: S->T, where S is the domain and T is the codomain.

2. What does it mean for a mapping to be bijective?

A bijective mapping is one that is both injective (one-to-one) and surjective (onto). This means that each element in the codomain is paired with exactly one element in the domain, and each element in the codomain has at least one corresponding element in the domain. In other words, every element in the codomain is mapped to by exactly one element in the domain, and every element in the codomain is mapped from by exactly one element in the domain.

3. How do you prove that a mapping is bijective?

To prove that a mapping is bijective, you must show that it is both injective and surjective. To prove injectivity, you must show that for every pair of distinct elements in the domain, their corresponding elements in the codomain are also distinct. To prove surjectivity, you must show that every element in the codomain has at least one corresponding element in the domain.

4. What is an inverse mapping?

An inverse mapping, denoted as f-1, is a mapping that undoes the effects of the original mapping f. In other words, if f maps an element x in the domain to an element y in the codomain, f-1 maps y back to x. This means that f-1 is the reverse of f.

5. Why does a bijective mapping need to have an inverse?

A bijective mapping must have an inverse in order to satisfy the definition of a bijective mapping. If a mapping is not bijective, it may not have an inverse, as there may not be a one-to-one correspondence between the elements of the domain and the elements of the codomain. Therefore, the existence of an inverse is a necessary condition for a mapping to be considered bijective.

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