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The map $(x, y) \mapsto (x, x y)$ is a diffeomorphism off the vertical axis

Example

Let $U = {(x, y) \in R^{2} : x \neq 0}$ . The map

$F : U \to U, F (x, y) = (x, x y),$

is a smooth diffeomorphism with inverse $G (u, v) = (u, v / u)$ . The same formula on all of $R^{2}$ omits every point $(0, v)$ with $v \neq 0$ .

Facts & Assumptions

Given: The map above.

[L1]

An injective regular $C^{k}$ map has open image and is a $C^{k}$ diffeomorphism onto that image (An injective regular $C^{k}$ map is a $C^{k}$ diffeomorphism onto its image).

[L2]

Finite componentwise products of $C^{k}$ Euclidean maps are $C^{k}$ ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition).

Verification

technique · direct

1.1

The coordinate functions are smooth, so [L2] makes $F (x, y) = (x, x y)$ smooth. For an increment $(h, k)$ , $F (x + h, y + k) - F (x, y) - (h, y h + x k) = (0, h k) .$ Since $2 ∣ h k ∣ \leq h^{2} + k^{2}$ , the remainder divided by $\sqrt{h^{2} + k^{2}}$ tends to zero. Hence $D F (x, y) = (\begin{array}{cc} 1 & 0 \\ y & x \end{array}), \det D F (x, y) = x .$ On $U$ , direct substitution gives $G (F (x, y)) = (x, y)$ and $F (G (u, v)) = (u, v)$ .

L2givenalgebra

2.1

Step 1.1 verifies the hypotheses and conclusion of [L1], so $F : U \to U$ is a smooth diffeomorphism. On the full plane, $F (0, y) = (0, 0)$ for every $y$ , and hence no point $(0, v)$ with $v \neq 0$ lies in its image.

step 1.1L1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, Example 8.5.3 (standard reference, not scraped)

The map (x,y)↦(x,xy)(x,y)↦(x,xy) is a diffeomorphism off the vertical axis