# Showing a mapping is onto and/or one to one

1. Feb 23, 2008

### FluxU

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.

2. Feb 23, 2008

### masnevets

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. Feb 23, 2008

### FluxU

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. Feb 24, 2008

### HallsofIvy

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. Feb 24, 2008

### jacobrhcp

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. Feb 25, 2008

### masnevets

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.