Mapping D* onto a Triangle: Investigating One-to-One Functions

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SUMMARY

The discussion focuses on the mapping function T defined on the domain D* = [0,1] x [0,1] as T(x*,y*) = (x*y*, x*). It concludes that T is not one-to-one due to the presence of the origin (0,0) in the image set. By analyzing parametric equations, participants determined that the image set D forms a triangle with vertices at (0,0), (t,t), and (0,t). The conversation emphasizes the importance of breaking down the function into simpler components for clarity.

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Homework Statement



Let D* = [0,1] x [0,1] and define T on D* by T(x*,y*) = (x*y*, x*). Determine the image set D. Is T one-to-one?

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The Attempt at a Solution



Okay... So I know it is not one to one, if you take out the point (x=0) then it is one-to-one, so you must be careful with the origin.
But I don't know how to find the image set D. Because there are two variables in the first component in T x and y*. How do I find D?
 
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hi der.physika! :smile:

it's easier if you break D* into two functions

what is the image of the function that sends (x, y) to (x, xy)?

(it may help to consider all the (x, 1)s)
 
tiny-tim said:
hi der.physika! :smile:

it's easier if you break D* into two functions

what is the image of the function that sends (x, y) to (x, xy)?

(it may help to consider all the (x, 1)s)


Okay... I think I found the solution

by using parametric equations

(t,0); (1,t); (t,1), (0,t) and plugging them in yields (0,0), (0,0), (t,t), (0,t) which makes it not 1 to 1. So... it gives me the triangle with those vertices. is this correct?
 
(just got up :zzz: …)
der.physika said:
… (t,0); (1,t); (t,1), (0,t) and plugging them in yields (0,0), (0,0), (t,t), (0,t) which makes it not 1 to 1. So... it gives me the triangle with those vertices. is this correct?

That's right! :smile:

(as you've seen, a short-cut that will work in any "non-folding" case is to simply find where all the corners go to :wink:)
 

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