Proof regarding 1-1, and onto mappings

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In summary, there exists a mapping from a set S to itself that is 1-1, but not onto if and only if there exists a mapping from S to itself that is onto, but not 1-1. This can be shown by defining a one-one and onto mapping f from S to a subset T of S, and then defining another mapping g from S to itself such that g(x) = f^-1(x) for x in T and g(x) = x for x not in T. The converse, however, remains to be proven and may involve representative points or equivalent points.
  • #1
calvino
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Show that there exists a mapping from a set S to itself that is 1-1, but not onto IFF there exists a mapping from a set S to itself that is onto, but not 1-1.

Firstly i show it, assuming that the one-one mapping (but not onto) exists.

Now i know that if there is a funtion that is 1-1, but not onto, to define a mapping f on the set (call it S), to a subset of S (call it T). This is such that f: S -> T is one-one and onto. Next, I define a mapping g:S->S such that

g(x) = { f^-1(x), when x is an element of T
{ x, whenever x is an element of S\T

That is my mapping that is then onto, but not one-one.


However, I'm stuch on the converse. That is, how do i prove, assuming a mapping from S to itself that is onto, but not one-one, implies that there is a mapping that is one-one, but not onto. ANY HELP>?
 
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  • #2
Does it have anything to do with representative points? (equivalent points).
 
  • #3


To prove the converse, we can use a similar approach. Let's assume that there exists a mapping g:S->S that is onto, but not one-one. This means that there exists at least one element in S that is mapped to by more than one element in S. Let's call this element y.

Now, we define a mapping h:S->S such that h(x) = y for all elements in S that are mapped to y by g. For all other elements in S, h(x) = x. This mapping is one-one, since no two elements in S are mapped to the same element in S. However, it is not onto, since there exists an element y in S that is not mapped to by any element in S.

Therefore, we have shown that the existence of a mapping from S to itself that is onto, but not one-one, implies the existence of a mapping that is one-one, but not onto. This completes the proof.
 

1. What is a one-to-one mapping?

A one-to-one mapping, also known as an injective function, is a function in which every element in the domain maps to a unique element in the range. This means that no two elements in the domain can map to the same element in the range.

2. How do you prove a function is one-to-one?

To prove that a function is one-to-one, you can use the horizontal line test. This involves drawing horizontal lines through the graph of the function. If each line intersects the graph at most once, then the function is one-to-one. Another way to prove one-to-one mapping is using the definition of one-to-one functions, which states that if f(a) = f(b), then a = b.

3. What is an onto mapping?

An onto mapping, also known as a surjective function, is a function in which every element in the range is mapped to by at least one element in the domain. This means that every element in the range has a preimage in the domain.

4. How do you prove a function is onto?

To prove that a function is onto, you can use the vertical line test. This involves drawing vertical lines through the graph of the function. If each line intersects the graph at least once, then the function is onto. Another way to prove onto mapping is using the definition of onto functions, which states that for every element in the range, there exists at least one element in the domain that maps to it.

5. What is the difference between one-to-one and onto mappings?

The main difference between one-to-one and onto mappings is that one-to-one mapping ensures that no two elements in the domain map to the same element in the range, while onto mapping ensures that every element in the range has a preimage in the domain. In other words, one-to-one mapping deals with unique mappings, while onto mapping deals with complete mappings.

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