Showing a mapping is onto and/or one to one

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In summary: Therefore, you only need to show that the map is injective to prove it is an isomorphism.In summary, to show that a mapping is one-to-one, you need to prove that if f(x) = f(y), then x = y, and in the case of a linear map, it is sufficient to show that f(x) = 0 only if x = 0. To show that a mapping is onto, you can either explicitly find a solution y for a given x, or take advantage of dimensions, with injectivity automatically implying surjectivity for two-dimensional vector spaces.
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FluxU
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Hello, I have a general question about isomorphisms of vector spaces. I understand the concepts of mappings being one to one, and onto, but how do I go about SHOWING that a mapping is either one to one, onto, one or the other or both?

For example, the mapping R^2-->R^2 defined by f(x,y) = (x-2y, x+y). I'm hoping someone can use this simple example to demonstrate how I would go about it. (it's bijective and an isomorphism btw).

The textbook we use doesn't show a general method for showing a mapping is 1:1/onto, but rather proves that a mapping ISN'T 1:1 or onto by means of a contradictory example.

Thanks in advance!
 
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  • #2
To show that a map is injective (one-to-one), you take some element x that maps to 0, and then prove that x has to be 0. So in your case, pick some x and y such that f(x,y) = 0.

Then x-2y=0 and x+y =0. Since x=-y, the first equation becomes -3y=0, and so x=y=0.
 
  • #3
That's exploiting the fact that the zero vector from one space maps to the zero vector in the target space, correct? That makes sense!

How about a method for testing for onto?
 
  • #4
Another way to prove one-to-one is to assume f(x)= f(y) and show that x= y, using whatever the definition of f is, of course. Proving that f(x)= 0 only if x= 0 works for linear transformations between vector spaces. If that is what you are working with, fine.

You prove a mapping is onto by showing that, for any y in the "target space" (range) the equation f(x)= y has at least one solution.
 
  • #5
and additionally, in the specific case the function f(x) is has values in R, it is surjective (onto) if the function is continuous and gives ±infinity as you take the limit of x to ± infinity.

in my experience, some classes you are not allowed to use that jet, so be careful.

moreover, any injective mapping between two finite sets of the same size is automatically onto (can you see why?)
 
  • #6
FluxU said:
That's exploiting the fact that the zero vector from one space maps to the zero vector in the target space, correct? That makes sense!

How about a method for testing for onto?

I am exploiting linearity, yes. As mentioned above, to check that a map is injective in general, one needs to show that if f(x) = f(y), then x=y. But in the case of a linear map, f(x) = f(y) implies f(x-y) = 0, so you only need to check the case that something maps to zero.

For onto, you can either explicitly find an x such that f(x) = y for a given y, or in your example, take advantage of dimensions, i.e., R^2 -> R^2 involves two-dimensional vector spaces, so injectivity automatically implies surjectivity.
 

1. How do you show that a mapping is onto?

To show that a mapping is onto, you need to prove that every element in the range of the mapping has at least one corresponding element in the domain. This can be done by taking an arbitrary element in the range and finding its preimage in the domain, or by showing that the range and the domain have the same cardinality.

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

Onto, or surjective, means that every element in the range of the mapping is mapped to by at least one element in the domain. One-to-one, or injective, means that every element in the range of the mapping is mapped to by at most one element in the domain. In other words, onto means that there are no elements in the range that are "left out", while one-to-one means that there are no elements in the range that are "doubled up".

3. Can a mapping be both onto and one-to-one?

Yes, a mapping can be both onto and one-to-one. Such a mapping is called a bijection. In this case, every element in the range of the mapping is mapped to by exactly one element in the domain, and every element in the domain has exactly one corresponding element in the range.

4. How do you prove that a mapping is one-to-one?

To prove that a mapping is one-to-one, you need to show that no two distinct elements in the domain are mapped to the same element in the range. This can be done by assuming that two elements in the domain have the same image, and then showing that this leads to a contradiction. Another approach is to show that the mapping is injective by using the definition of one-to-one functions.

5. What is a counterexample for a mapping that is onto but not one-to-one?

A counterexample for a mapping that is onto but not one-to-one is a mapping that has two elements in the domain that are mapped to the same element in the range. For example, the mapping f: {1,2,3} → {1,2} defined by f(1) = 1, f(2) = 2, f(3) = 1 is onto, but not one-to-one, since 1 and 3 in the domain are both mapped to 1 in the range.

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