Find an open set V containing ( 1 , 1 ) such that f | V : V → f ( V ) is bijective, where f : R 2 → R 2 &amp; f ( x , y ) = ( x y , x − y )My Question: Clearly f ( x 1 , y 1 ) = f ( x 2 , y 2 ) ⟹ x 1 y 1 = x 2 y 2 &amp; x 1 − y 1 = x 2 − y 2 , thus x 1 2 + y 1 2 = x 2 2 + y 2 2 . Hence both ( x 1 , y 1 ) and ( x 2 , y 2 ) are points of intersection of the same line and circle. But Then how do I find the open set V?

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Find an open set V containing   (  1  ,  1  ) such that   f            |        V    :  V  →  f  (  V  ) is bijective, where   f  :            R        2    →            R        2      &amp;amp;    f  (  x  ,  y  )  =  (  x  y  ,  x  −  y  )My Question: Clearly   f  (      x    1    ,      y    1    )  =  f  (      x    2    ,      y    2    )    ⟹        x    1        y    1    =      x    2        y    2      &amp;amp;        x    1    −      y    1    =      x    2    −      y    2  , thus       x    1    2    +      y    1    2    =      x    2    2    +      y    2    2  . Hence both   (      x    1    ,      y    1    ) and   (      x    2    ,      y    2    ) are points of intersection of the same line and circle. But Then how do I find the open set V?

dobradisamgn · Accepted Answer

There is no such open set! Indeed,   f  (  t  ,  t  )  =  (      t          2        ,  0  ) and   f  (  −  t  ,  −  t  )  =  (      t          2        ,  0  ). Any open set containing (0,0) contains   (  t  ,  t  ) and   (  −  t  ,  t  ) for some   t  ≠  0 so f is not injective an any such open set.Answer for the edited version:We have   (      x    1    +      y    1        )          2        =  (      x    1    −      y    1        )          2        +  4      x    1        y    1    =  (      x    2    −      y    2        )          2        +  4      x    2        y    2    =  (      x    2    +      y    2        )          2       which gives       x    1    +      y    1    =  ±  (      x    2    +      y    2    ). If we choose any neighborhood of (1,1) in which the coordinates are both positive then we get       x    1    +      y    1    =      x    2    +      y    2   Since we also have       x    1    −      y    1    =      x    2    −      y    2   we see that       x    1    =      y    1   and       x    2    =      y    2   so f is injective in such a neighborhood.

Find an open set V containing (1,1) such that f|_V:V -> f(V) is bijective, where f:RR^2 -> RR^2 & f(x,y)=(xy,x−y)

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