Let \[ W=\{(u,v)\in\mathbb R^2:u>v\} \] and define \[ T(u,v)=(u+v,u^2+v^2). \]
- Prove that \(T\) is locally one-to-one.
- Determine the range of \(T\) and prove that \(T\) is globally one-to-one.
Solution
1. The Jacobian determinant of \(T\) at \((u,v)\in W\) is\[ \det DT(u,v)=2(v-u)\neq0. \]
The derivative matrix is \[ DT(u,v) = \begin{pmatrix} 1&1\\ 2u&2v \end{pmatrix}, \] so \[ \det DT(u,v) = 2v-2u = 2(v-u). \] Since \((u,v)\in W\) means \(u>v\), this determinant is nonzero.
2. The map \(T\) is locally one-to-one on \(W\).
By step 1, the derivative of \(T\) is invertible at every point of the open set \(W\). The inverse function theorem therefore gives, around each point of \(W\), a neighborhood on which \(T\) is one-to-one.
3. If\[ T(u,v)=(s,q), \]then\[ q>\frac{s^2}{2}. \]
Set \[ s=u+v, \qquad d=u-v. \] Since \((u,v)\in W\), one has \(d>0\). Moreover, \[ \begin{aligned} 2q &= 2u^2+2v^2\\ &= (u+v)^2+(u-v)^2\\ &= s^2+d^2. \end{aligned} \] Thus \[ q = \frac{s^2+d^2}{2} > \frac{s^2}{2}. \]
4. Conversely, if \((s,q)\in\RR^2\) satisfies\[ q>\frac{s^2}{2}, \]then it has a preimage in \(W\), namely\[ u = \frac{s+\sqrt{2q-s^2}}{2}, \qquad v = \frac{s-\sqrt{2q-s^2}}{2}. \]
The hypothesis makes \[ d\coloneqq\sqrt{2q-s^2} \] a positive real number. The displayed definitions give \[ u+v=s \qquad\text{and}\qquad u-v=d>0, \] so \((u,v)\in W\). Also, \[ u^2+v^2 = \frac{(u+v)^2+(u-v)^2}{2} = \frac{s^2+d^2}{2} =q. \] Hence \(T(u,v)=(s,q)\).
5. The range of \(T\) is\[ \boxed{ T(W) = \left\{ (s,q)\in\RR^2: q>\frac{s^2}{2} \right\} }. \]
Step 3 gives the inclusion from left to right, and step 4 gives the reverse inclusion.
6. The map \(T\) is globally one-to-one on \(W\).
Suppose \[ T(u,v)=(s,q) \] with \((u,v)\in W\). Step 3 gives \[ (u-v)^2 = 2q-s^2. \] Because \(u-v>0\), necessarily \[ u-v=\sqrt{2q-s^2}. \] Together with \(u+v=s\), this uniquely determines \[ u = \frac{s+\sqrt{2q-s^2}}2, \qquad v = \frac{s-\sqrt{2q-s^2}}2. \] Thus every point in the range has at most one preimage.
7. Q.E.D.
Step 2 proves local one-to-one behavior, step 5 determines the range, and step 6 proves global injectivity.