# One-to-One Mapping

1. Oct 25, 2009

### CYRANEX

Hi

My book defines one-to-one mapping as

A mapping T is one-to-one on D* if for (u,v) and (u',v') ∈ D*,
T(u,v) = T(u', v') implies that u = u' and v = v'

I don't really understand what they are trying to say, because right now what I'm getting from this information is that only functions that are equal to their derivative can be mapped one-to-one, but doesn't that mean only lines can be one to one functions?

Another thing
I know the Jacobian ≠ 0 has something to do with a function having an inverse and being mapped one-to-one, but my book just skips over that, so could someone please explain that. Also Jacobian Determinant is the determinant of the derivative matrix what does that have anything to do with inverse functions, and moreso what does a determinant even mean.

Help, I really want to learn mathematics and physics, but I always get bogged in technicalities and poorly written text books :(

2. Oct 25, 2009

### emyt

that isn't the notation for differentiation, just stating that v' is not v.

a one-to-one mapping has a unique mapping for each unique x

3. Oct 26, 2009

### mikeph

that statement just means that any point T has only one point on the domain which maps to it- ie. if it has two points T(x) and T(y) that are the same then x = y, so it's really just the same point.

4. Oct 26, 2009

### CYRANEX

Thanks guys you helped a bit, but I'm still confused
are there better definitions for one-to-one mapping?

@emyt
Ok, that makes sense that v' isn't the derivative of v, but I still have no idea why v = v' and u = u'

@Mikey W
I am not really sure what you are trying to say, but from what I gather your saying is that if x = y, then T(x) = T(y). And if T(x) = e and T(y) = f then e = f. That doesn't explain how a function would be one to one if it satisfied the book definition:

5. Oct 26, 2009

### emyt

a one to one mapping has a unique "output" for each "input", so if the mapping T is one to one, and T(u,v) = T(u',v'), then u =u' and v = v' - or else it wouldn't have been a one-to-one map.

Last edited: Oct 26, 2009
6. Oct 26, 2009

### CYRANEX

Ok I get it now :)
Thanks so much
The book is simply stating that
if T(u,v) = T(u',v')
then u must equal u'
and v must equal v'
or else it is not a one to one map

But that's what you guys have been saying...

I still don't understand why the Jacobian determinant cannot equal zero though
Jacobian determinant is this
$$\frac{\partial{x}}{\partial{u}} \frac{\partial{y}}{\partial{v}} - \frac{\partial{x}}{\partial{v}} \frac{\partial{y}}{\partial{u}}$$

I know that if it doesn't equal zero that means that
$$\frac{\partial{x}}{\partial{u}} \frac{\partial{y}}{\partial{v}} \neq \frac{\partial{x}}{\partial{v}} \frac{\partial{y}}{\partial{u}}$$

but what does the product of partials mean?

7. Oct 26, 2009

### HallsofIvy

The Jacobian is the determinant of the matrix representing the "linear map" that approximates the function locally (think of it as the tangent plane to the surface). A linear map, from a vector space to itself, is one-to-one if and only if it is invertible and that is true if and only if its determinant is non-zero.